3 Marks Question - Maths STD 9 Questions - Vidyadip

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Question 13 Marks

Simplify the following products:
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$

Answer

In the given problem, we have to find product of
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
taking x as common factor $\text{x}(\text{x}^2 -3\text{x}-1)(\text{x}^2-3\text{x} + 1)$
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)\\=\Big[\text{x}\big(\text{x}^2-3\text{x}-1\big)\big(\text{x}^2-3\text{x}+1\big)\Big]$
$=\text{x}\Big[\big\{\big(\text{x}^2-3\text{x}\big)-1\big\}\big\{\big(\text{x}^2-3\text{x}\big)+1\big\}\Big]$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$
$\big(\text{x}^3-3\text{x}^2-\text{x}\big)\big(\text{x}^2-3\text{x}+1\big)\\=\text{x}\Big[\big(\text{x}^2-3\text{x}\big)^2-1^2\Big]$
$=\text{x}\big(\text{x}^4-6\text{x}^3+9\text{x}^2-1\big)$
$=\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}$
Hence the value of $(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$ is $\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}.$

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Question 23 Marks

Simplify the following products:
$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$

Answer

In the given problem, we have to find product of $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$
On rearranging we get
$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)\\=\big[\text{x}^2+(\text{x}-2)\big]\big[\text{x}^2-(\text{x}-2)\big]$'
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$
$(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)\\=\Big[+(\text{x}^2)^2-(\text{x}-2)^2\Big]$
$=\text{x}^4-\big(\text{x}^2-2\times2\times\text{x}+2^2\big)$
$=\text{x}^4-\big(\text{x}^2-4\text{x}+4\big)$
$=\text{x}^4-\text{x}^2+4\text{x}-4$
Hence the value of $(\text{x}^2 +\text{x}- 2)(\text{x}^2-\text{x} + 2)$ is $\text{x}^4-\text{x}^2+4\text{x}-4.$

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Question 33 Marks

Simplify the following products:
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$

Answer

In the given problem, we have to find product of
$(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$
On rearranging we get $\Big(\big[2\text{x}^4-4\text{x}^2\big]+1\Big)\Big(\big[2\text{x}^4-4\text{x}^2\big]-1\Big)$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y)}=\text{x}^2-\text{y}^2$
$\big(2\text{x}^4-4\text{x}^2+1\big)\big(2\text{x}^4-4\text{x}^2-1\big)\\=\big[2\text{x}^4-4\text{x}^2\big]^2-1^2$
$=\big[4\text{x}^8+16\text{x}^4-2\times2\text{x}^4\times4\text{x}^2-1\big]$
$=4\text{x}^8+16\text{x}^4-16\text{x}^6-1$
Hence the value of $(2\text{x}^4 -4\text{x}^2+1)(2\text{x}^4-4\text{x}^2-1)$ is $4\text{x}^8+16\text{x}^4-16\text{x}^6-1$

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Question 43 Marks

If x = 3, find the values of the following using in identity:
$\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\Big(\frac{\text{x}^2}{9}+\frac{9}{\text{x}^2}+1\Big)$

Answer

We have,
$\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\Big(\frac{\text{x}^2}{9}+\frac{9}{\text{x}^2}+1\Big)$
$=\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\bigg[\Big(\frac{\text{x}}{3}\Big)+\Big(\frac{3}{\text{x}}\Big)^2+\frac{3}{\text{x}}\times\frac{\text{x}}{3}\bigg]$
$=\Big(\frac{3}{\text{x}}\Big)^3-\Big(\frac{\text{x}}{3}\Big)^3$ $\big[\because\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{y}^2+\text{xy})\big]$
$=\frac{27}{\text{x}^3}-\frac{\text{x}^3}{27}$
$=\frac{27}{(3)^3}-\frac{(3)^3}{27}$ $\big[\because\text{x}=3\big]$
$=\frac{27}{27}-\frac{27}{27}$
$=1-1$
$=0$
$\therefore\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\Big(\frac{\text{x}^2}{9}+\frac{9}{\text{x}^2}+1\Big)=0$

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Question 53 Marks

If x = 3 and y = -1, find the values of the following using in identity:
$\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{49}+\frac{\text{y}^2}{9}-\frac{\text{xy}}{21}\Big)$

Answer

We have,
$\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{49}+\frac{\text{y}^2}{9}-\frac{\text{xy}}{21}\Big)$
$=\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\bigg[\Big(\frac{\text{x}}{7}\Big)^2+\Big(\frac{\text{y}}{3}\Big)-\frac{\text{x}}{7}\times\frac{\text{y}}{3}\bigg]$
$=\Big(\frac{\text{x}}{7}\Big)^3+\Big(\frac{\text{y}}{3}\Big)^3$ $\big[\because\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{\text{x}^3}{343}+\frac{\text{y}^3}{27}$
$=\frac{(3)^3}{343}+\frac{(-1)^3}{27}$ $\big[\because\text{x}=3\ \text{and}\ \text{y}=-1\big]$
$=\frac{27}{343}+\frac{-1}{27}$
$=\frac{729-343}{9261}=\frac{386}{9261}$
$\therefore\Big(\frac{\text{x}}{7}+\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{49}+\frac{\text{y}^2}{9}-\frac{\text{xy}}{21}\Big)=\frac{386}{9261}$

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Question 63 Marks

If x = 3 and y = -1, find the values of the following using in identity:
$\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)$

Answer

We have,

$\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)$

$=\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\bigg[\Big(\frac{\text{x}}{4}\Big)^2+\frac{\text{x}}{4}\times\frac{\text{y}}{3}+\Big(\frac{\text{y}}{3}\Big)^2\bigg]$

$=\Big(\frac{\text{x}}{4}\Big)^3-\Big(\frac{\text{y}}{3}\Big)^3$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big]$

$=\frac{\text{x}^3}{64}-\frac{\text{y}^3}{27}$

$=\frac{(3)^3}{64}-\frac{(-1)^3}{27}$ $\big[\because\text{x}=3,\ \text{y}=-1\big]$

$=\frac{27}{64}+\frac{1}{27}$

$=\frac{729+64}{1728}=\frac{793}{1728}$

$\therefore\Big(\frac{\text{x}}{4}-\frac{\text{y}}{3}\Big)\Big(\frac{\text{x}^2}{16}+\frac{\text{xy}}{12}+\frac{\text{y}^2}{9}\Big)=\frac{793}{1728}$

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Question 73 Marks

If x = 3 and y = -1, find the values of the following using in identity:
$\Big(\frac{5}{\text{x}}+5\text{x}\Big)\Big(\frac{25}{\text{x}^2}-25+25\text{x}^2\Big)$

Answer

We have,
$\Big(\frac{5}{\text{x}}+5\text{x}\Big)\Big(\frac{25}{\text{x}^2}-25+25\text{x}^2\Big)$
$=\Big(\frac{5}{\text{x}}+5\text{x}\Big)\bigg[\Big(\frac{5}{\text{x}}\Big)^2-\frac{5}{\text{x}}\times5\text{x}+(5\text{x})^2\bigg]$
$=\Big(\frac{5}{\text{x}}\Big)^3+(5\text{x})^3$ $\big[\because\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{125}{\text{x}^3}+125\text{x}^3$
$=\frac{125}{(3)^3}+125(3)^3$ $\big[\because\text{x}=3\big]$
$=\frac{125}{27}+125\times27$
$=\frac{125}{27}+3375$
$=\frac{125+3375\times27}{27}=\frac{125+91125}{27}$
$=\frac{91250}{27}$

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Question 83 Marks

If x = 3 and y = -1, find the values of the following using in identity:
(9y² - 4x²)(81y⁴ + 36x²y² + 16x⁴)

Answer

We have,
(9y² - 4x²)(81y⁴ + 36x²y² + 16x⁴)
= (9y² - 4x²)$\big[$(9y²)² + 9y² × 4x² + (4x²)²$\big]$
= (9y²)³ - (4x²)³ $\big[\because$ a³ - b³ = (a - b)(a² + ab + b²)$\big]$
= 729y⁶ - 64x⁶
= 729 × (-1)⁶ - 64(3)⁶ $\big[\because$ x = 3 and y = -1$\big]$
= 729 - 64 × 729
= 729 - 46656
= -45927
(9y² - 4x²)(81y⁴ + 36x²y² + 16x⁴) = -45927

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Question 93 Marks

If x = -2 and y = 1, by using an identity find the value of the following:
$\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\Big(\frac{4}{\text{x}^2}+\frac{\text{x}^2}{4}+1\Big)$

Answer

We have,
$\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\Big(\frac{4}{\text{x}^2}+\frac{\text{x}^2}{4}+1\Big)$
$=\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\bigg[\Big(\frac{2}{\text{x}}\Big)^2+\Big(\frac{\text{x}}{2}\Big)+\frac{2}{\text{x}}\times\frac{\text{x}}{2}\bigg]$
$=\Big(\frac{2}{\text{x}}\Big)^3-\Big(\frac{\text{x}}{2}\Big)^3$ $\big[\therefore\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{b}^2+2\text{ab})\big]$
$=\frac{8}{\text{x}^3}-\frac{\text{x}^3}{8}$
$=\frac{8}{(-2)^3}-\frac{(-2)^3}{8}$ $\big[\therefore\text{x}=-2\big]$
$=\frac{8}{-8}+\frac{8}{8}$
$=-1+1=0$
$\therefore\Big(\frac{2}{\text{x}}-\frac{\text{x}}{2}\Big)\Big(\frac{4}{\text{x}^2}+\frac{\text{x}^2}{4}+1\Big)=0$

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Question 103 Marks

If x = -2 and y = 1, by using an identity find the value of the following:
$\Big(5\text{y}+\frac{15}{\text{y}}\Big)\Big(25\text{y}^2-75+\frac{225}{\text{y}^2}\Big)$

Answer

We have,

$\Big(5\text{y}+\frac{15}{\text{y}}\Big)\Big(25\text{y}^2-75+\frac{225}{\text{y}^2}\Big)$

$\Big(5\text{y}+\frac{15}{\text{y}}\Big)\bigg[\Big(5\text{y})^2-5\text{y}\times\frac{15}{\text{y}}\Big(\frac{15}{\text{y}}\Big)^2\bigg]$

$=(5\text{y})^3+\Big(\frac{15}{\text{y}}\Big)^3$ $\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$

$=125\text{y}^3+\frac{3375}{\text{y}^3}$

$=125(1)^3+\frac{3375}{(1)^3}$ $\big[\therefore\text{y}=1\big]$

$=125+3375$

$=3500$

$\therefore\Big(5\text{y}+\frac{15}{\text{y}}\Big)\Big(25\text{y}^2-75+\frac{225}{\text{y}^2}\Big)=3500$

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Question 113 Marks

If x = -2 and y = 1, by using an identity find the value of the following:
(4y² - 9x²)(16y⁴ + 36x²y² + 81x⁴)

Answer

We have,
(4y² - 9x²)(16y⁴ + 36x²y² + 81x⁴)
= (4y² - 9x²)$\big[$(4y²)² + 4y² × 8x² + (9x²)²$\big]$
= (4y²)³ - (9x²)³ $\big[\because$ a³ - b³ = (a - b)(a² + ab + b²)$\big]$
= 64y⁶ - 729x⁶
= 64(1)⁶ - 729(-2)⁶ $\big[\because$ x = 2 and y = 1$\big]$
= 64 - 729 × 64
= 64 - 46656
= -465992
$\therefore$ (4y² - 9x²)(16y⁴ + 36x²y² + 81x⁴) = -465992

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Question 123 Marks

If $\text{x}^2+\frac{1}{\text{x}^2}=66,$ find the value of $\text{x}-\frac{1}{\text{x}}.$

Answer

We have,
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=66-2$ $\big[\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=66\big]$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=64$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=(\pm8)^2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)=\pm8$

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Question 133 Marks

If a² + b² + c² = 16 and ab + bc + ca = 10, find the value of a + b + c.

Answer

We know that,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(x + y + z)² = 16 + 2(10)
(x + y + z)² = 36
$\big(\text{x}+\text{y}+\text{z}\big) = \sqrt{36}$
$\big(\text{x}+\text{y}+\text{z}\big) =\pm6$
Hence, value of required expression $\big(\text{a}+\text{b}+\text{c}) =\pm6$

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Question 143 Marks

If a + b + c = 9 and ab + bc + ca = 23, find the value of a² + b² + c².

Answer

We know that,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
9² = a² + b² + c² + 2(23)
81 = a² + b² + c² + 46
a² + b² + c² = 81 - 46
a² + b² + c² = 35
Hence, value of required expression a² + b² + c² = 35

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Question 153 Marks

If a + b + c = 0, then write the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}.$

Answer

We have to find the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Given a + b + c = 0
Using identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
Put a + b + c = 0
a³ + b³ + c³ - 3abc = (0)(a² + b² + c² - ab - bc - ca)
a³ + b³ + c³ - 3abc = 0
a³ + b³ + c³ = 3abc
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{abc}}+\frac{\text{c}^3}{\text{abc}}=3$
$\frac{\text{a}\times\text{a}\times\text{a}}{\text{abc}}+\frac{\text{b}\times\text{b}\times\text{b}}{\text{abc}}+\frac{\text{c}\times\text{c}\times\text{c}}{\text{abc}}=3$
$\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Hence the value of $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$ is 3.

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Question 163 Marks

If a + b + c = 0 and a² + b² + c² = 16, find the value of ab + bc + ca.

Answer

We know that,
$\big[\therefore$(a + b + c)² = a²+ b²+ c² + 2ab + 2bc + 2ca$\big]$
(0)² = 16 + 2(ab + bc + ca)
2(ab + bc + ca)= -16
ab + bc + ca = -8
Hence, value of required express ab + bc + ca = -8

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Question 173 Marks

If a + b = 8 and ab = 6, find the value of a³ + b³.

Answer

We have,
a³ + b³ = (a + b)(a² - ab + b²)
= (a + b)(a² + b² - ab)
= (a + b)(a² + b² - ab + 2ab - 2ab) [Adding and substracting 2ab in the second break]
= (a + b)$\big[$(a² + b² + 2ab) - 3ab$\big]$
= (a + b)$\big[$(a + b)² - 3ab$\big]$ $\big[\because$ (a + b)² = a² + b² + 2ab$\big]$
= 8 × $\big[$(8)² - 3 × 6$\big]$ $\big[\because$ a + b = 8 and ab = 6$\big]$
= 8 × [64 - 18]
= 8 × 46
= 368
$\therefore$ a³ + b³ = 368

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Question 183 Marks

If a + b = 7 and ab = 12, find the value of a² + b²

Answer

We have to find the value of a² + b²
Given a + b = 7, ab = 12
Using identity (a + b)² = a² + 2ab + b²
By substituting the value of a + b = 7, ab = 12 we get
(a + b)² = a² + b² + 2 × ab
(7)² = a² + b² + 2 × 12
49 = a² + b² + 24
By transposing +24 to left hand side we get ,
49 - 24 = a² + b²
25=a² + b²
Hence the value of a² + b² is 25.

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Question 193 Marks

If a - b = 6 and ab = 20, find the value of a³ - b³.

Answer

We have,
a³ - b³ = (a - b)(a² + ab + b²)
= (a - b)(a² + ab + b²- 2ab + 2ab) [Adding and substracting 2ab in the second break]
= (a - b)$\big[$(a² + b² - 2ab) + 3ab$\big]$
= (a - b)$\big[$(a - b)² + 3ab$\big]$ $\big[\because$ (a - b)² = a² + b² - 2ab$\big]$
= 6 × $\big[$(6)² + 3 × 20$\big]$ $\big[\because$ a - b = 6 and ab = 20$\big]$
= 6 × [36 + 60]
= 6 × 96
= 576
$\therefore$ a³ - b³ = 576

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Question 203 Marks

If a - b = 4 and ab = 21, find the value of a³ - b³.

Answer

In the given problem, we have to find the value of a³ - b³
Given a - b = 4, ab = 21
We shall use the identity (a - b)³ = a³ - b³ - 3ab(a - b)
Here putting a - b = 4, ab = 21,
⇒ (4)³ = a³ - b³ + 3(21)(4)
⇒ 64 = a³ - b³ - 252
⇒ 64 – 252 = a³ - b³
⇒ 316 = a³- b³
Hence, the value of a³- b³ is 316.

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Question 213 Marks

If a + b = 10 and ab = 21, find the value of a³ + b³.

Answer

Given,
a + b = 10, ab = 21
We know that, (a + b)³ = a³ + b³ + 3ab(a + b) ...1
Substitute a + b = 10, ab = 21 in eq. 1
⇒ (10)³ = a³ + b³ + 3(21)(10)
⇒ 1000 = a³ + b³ + 630
⇒ 1000 – 630 = a³ + b³
⇒ 370 = a³+ b³
Hence, the value of a³+ b³ = 370

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Question 223 Marks

If 9x²+ 25y²= 181 and xy = -6, find the value of 3x + 5y.

Answer

We have,
(3x + 5y)² = (3x)² + (5y)² + 2 × 3x × 5y
⇒ (3x + 5y)² = 9×² + 25y² + 30xy
= 181 + 30(-6) [Since, 9x²+ 25y²= 181 and xy = - 6]
⇒ (3x + 5y)² = 1
⇒ (3x + 5y)² = $ (\pm 1)2$
⇒ 3x + 5y = $\pm1$

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Question 233 Marks

If 3x - 2y = 11 and xy = 12, find the value of 27x³ - 8y³.

Answer

Given, 3x - 2y = 11, xy = 12
We know that (a – b)³ = a³ - b³ - 3ab(a + b)
(3x - 2y)³ = 11³
⇒ 27x³ - 8y³- (18 × 12 × 11) = 1331
⇒ 27x³ - 8y³- 2376 = 1331
⇒ 27x³ - 8y³= 1331 + 2376
⇒ 27x³ - 8y³= 3707
Hence, the value of 27x³ - 8y³= 3707

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Question 243 Marks

Find the value of 4x² + y² + 25z² + 4xy - 10yz - 20zx when x = 4, y = 3 and z = 2.

Answer

4x² + y² + 25z² + 4xy - 10yz - 20zx
= (2x)² + y² + (-5z)² + 2(2x)(y) + 2(y)(-5z) + 2(-5z)(2x)
= (2x + y - 5z)²
= (2(4) + 3 - 5(2))²
= (8 + 3 - 10)²
= (1)²
= 1
Hence value of the equation is equals to 1

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Question 253 Marks

Find the following products:
(4x - 3y + 2z)(16x² + 9y² + 4z² + 12xy + 6yz - 8zx)

Answer

We have,
(4x - 3y + 2z)(16x² + 9y² + 4z² + 12xy + 6yz - 8zx)
= (4x +(-3y) + 2z)$\big[$(4x)² + (-3y)² + (2z)² - (4x)(-3y) - (-3y)(2z) - (2z)(4x)$\big]$
= (4x)³ + (-3y)³ + (2z)³ - 3 × (4x) × (-3y)(2z) $\big[\because$ a³ + b³ + c³ = 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)$\big]$
= 64x³ - 27y³ + 8z³ + 72xyz
$\therefore$ (4x - 3y + 2z)(16x² + 9y² + 4z² + 12xy + 6yz - 8zx) = 64x³ - 27y³ + 8z³ + 72xyz

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Question 263 Marks

Find the following products:
(3x - 4y + 5z)(9x² + 16y² + 25z² + 12xy - 15zx + 20yz)

Answer

We have,
(3x - 4y + 5z)(9x² + 16y² + 25z² + 12xy - 15zx + 20yz)
= (3x + (-4y) + 5z)$\big($(3x)2 + (-4y)2 + (5z)2 - (3x)(-4y) - (5z)(3x)$\big)$
= (3x)³ + (-4y)³ + (5z)³ - 3(3x)(-4y)(5z)
$\big[\because$ a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)$\big]$
= 27x³ - 64y³ + 125z³ + 180xyz
$\therefore$ (3x - 4y + 5z)(9x² + 16y² + 25z² + 12xy - 15zx + 20yz) = 27x³ - 64y³ + 125z³ + 180xyz

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Question 273 Marks

Find the following products:
(3x + 2y + 2z)(9x² + 4y² + 4z² - 6xy - 4yz - 6zx)

Answer

We have,
(3x + 2y + 2z)(9x² + 4y² + 4z² - 6xy - 4yz - 6zx)
= (3x + 2y + 2z)$\big($(3x)² + (2y)² + (2z)² - 3x × 2y × 2z - 2z × 3x$\big)$
= (3x)³ + (2y)³ + (2z)³ - 3 × 3x × 2y × 2z $\big[\because$ a³ + b³ + c³ = 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)$\big]$
= 27x³ + 8y³ + 8z³ - 36xyz
$\therefore$ (3x + 2y + 2z)(9x² + 4y² + 4z² - 6xy - 4yz - 6zx) = 27x³ + 8y³ + 8z³ - 36xyz

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Question 283 Marks

Find the following products:
(2a - 3b - 2c)(4a² + 9b² + 4c² + 6ab - 6bc + 4ca)

Answer

We have,
(2a - 3b - 2c)(4a² + 9b² + 4c² + 6ab - 6bc + 4ca)
= (2a + (-3b) + (-2c)) + $\big($(2a)² + (-3b)² + (-2c)² - (2a)(-3b)(-2c) - (-2c)(2a)$\big)$
= (2a)³ + (-3b)³ + (-2c)³ - 3(2a)(-3b)(-2c)
$\big[\because$ a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)$\big]$
= 8a³ - 27b³ - 8c³ - 36abc
$\therefore$ (2a - 3b - 2c)(4a² + 9b² + 4c² + 6ab - 6bc + 4ca) = 8a³ - 27b³ - 8c³ - 36abc

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Question 293 Marks

Evaluate the following using identities:
991 × 1009

Answer

We have,
991 × 1009
= (1000 - 9)(1000 + 9)
= (1000)² - (9) [(a + b)(a - b) = a² - b²]
= 1000000 - 81 [Where a = 1000 and b = 9]
= 999919
Therefore, 991 × 1009 = 999919

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Question 303 Marks

Evaluate the following using identities:
117 × 83

Answer

We have,
117 × 83
= (100 + 17)(100 - 17)
= (100)² - (17)²[(a + b)(a - b) = a² - b²]
= 10000 - 289 [Where a = 100 and b = 17]
= 9711
Therefore, 117 × 83 = 9711

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Question 313 Marks

Evaluate the following:
(99)³

Answer

We know that (a - b)³ = a³ - b³ - 3ab(a - b)
⇒ (99)³ can be written as (100 - 1)³
Here, a = 100 and b = 1
(99)³= (100 - 1)³
= (100)³ - (1)³ - 3(100)(1)(100 - 1)
= 1000000 - 1 - (300 × 99)
= 1000000 - 1 - 29700
= 1000000 - 29701
= 970299
The value of (99)³ = 970299

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Question 323 Marks

Evaluate the following:
(9.9)³

Answer

We know that (a - b)³ = a³ - b³ - 3ab(a - b)
⇒ (9.9)³ can be written as (10 - 0.1)³
Here, a = 10 and b = 0.1
(9.9)³ = (10 - 0.1)³
= (10)³ - (0.1)³ - 3(10)(0.1)(10 - 0.1)
= 1000 - 0.001 - (3 × 9.9)
= 1000 - 0.001 - 29.7
= 1000 - 29.701
= 970.299
The value of (9.9)³ = 970.299

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Question 333 Marks

Evaluate the following:
(98)³

Answer

We know that (a - b)³ = a³ - b³ - 3ab(a - b)
⇒ (98)³ can be written as (100 - 2)³
Here, a = 100 and b = 2
(98)³ = (100 - 2)³
= (100)³ - (2)³ - 3(100)(2)(100 - 2)
= 1000000 - 8 - (600 × 102)
= 1000000 - 8 - 58800
= 941192
The value of (98)³ = 941192

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Question 343 Marks

Evaluate the following:
93³ - 107³

Answer

Given,
93³ - 107³
the above equation can be written as (100 - 7)³ - (100 + 7)³
we know that, (a - b)³ - (a + b)³ = -2[b³ + 3ba²]
here, a = 93, b = 107
(100 - 7)³ - (100 + 7)³= -2[7³ + 3(100)²(7)]
= -2[343 + 210000]
= -2[210343]
= -420686
The value of 93³ - 107³ = -420686

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Question 353 Marks

Evaluate the following:
(598)³

Answer

We know that (a - b)³ = a³ - b³ - 3ab(a - b)
⇒ (598)³ can be written as (600 - 2)³
Here, a = 600 and b = 2
(598)³ = (600 - 2)³
= (600)³ - (2)³ - 3(600)(2)(600 - 2)
= 216000000 - 8 - (3600 × 598)
= 216000000 - 8 - 2152800
= 216000000 - 2152808
= 213847192
The value of (598)³ = 213847192

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Question 363 Marks

Evaluate the following:
46³ + 34³

Answer

Given,
46³ + 34³
The above equation can be written as (40 + 6)³ + (40 - 6)³
We know that, (a + b)³ + (a - b)³ = 2[a³ + 3ab²]
here, a = 40, b = 4
(40 + 6)³ + (40 - 6)³= 2[40³ + 3(6)²(40)]
= 2[64000 + 4320]
= 2[68320]
= 1366340
The value of 46³ + 34³ = 1366340

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Question 373 Marks

Evaluate the following:
111³ - 89³

Answer

Given,
111³ - 89³
the above equation can be written as (100 + 11)³ - (100 - 11)³
we know that, (a + b)³ - (a - b)³ = 2[b³ + 3ab²]
here, a = 100 b = 11
(100 + 11)³ - (100 - 11)³= 2[11³ + 3(100)²(11)]
= 2[1331 + 330000]
= 2[331331]
= 662662
The value of 111³ - 89³ = 662662

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Question 383 Marks

Evaluate the following:
(10.4)³

Answer

We know that (a + b)³ = a³ + b³ + 3ab(a + b)
⇒ (10.4)³ can be written as (10 + 0.4)³
Here, a = 10 and b = 0.4
(10.4)³ = (10 + 0.4)³
= (10)³ + (0.4)³ + 3(10)(0.4)(10 + 0.4)
= 1000 + 0.064 + (12 × 10.4)
= 1000 + 0.064 + 124.8
= 1000 + 124.864
= 1124.864
The value of (10.4)³ = 1124.864

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Question 393 Marks

Evaluate the following:
104³ + 96³

Answer

Given,
104³ + 96³
the above equation can be written as (100 + 4)³ + (100 - 4)³
we know that, (a + b)³ + (a - b)³ = 2[a³ + 3ab²]
here, a= 100, b = 4
(100 + 4)³ - (100 - 4)³= 2[100³ + 3(4)²(100)]
= 2[1000000 + 4800]
= 2[1004800]
= 200960
The value of 104³ + 96³ = 2009600

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Question 403 Marks

Evaluate the following:
(103)³

Answer

We know that (a + b)³ = a³ + b³ + 3ab(a + b)
⇒ (103)³ can be written as (100 + 3)³
Here, a = 100 and b = 3
(103)³ = (100 + 3)³
= (100)³ + (3)³ + 3(100)(3)(100 + 3)
= 1000000 + 27 + (900 × 103)
= 1000000 + 27 + 92700
= 1092727
The value of (103)³ = 1092727

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Question 413 Marks

Evaluate:
$\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3=\Big(\frac{5}{6}\Big)^3$

Answer

Let $\text{a}=\frac{1}{2},\ \text{b}=\frac{1}{3}$ and $\text{c}=\frac{-5}{6}$
Then,
$\text{a} + \text{b} + \text{c}=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$
$=\frac{3+2}{6}-\frac{5}{6}$
$\Rightarrow\text{a}+\text{b}+\text{c}=\frac{5}{6}-\frac{5}{6}=0$
$\therefore\text{a}^3+\text{b}^3+\text{c}^3=3\text{abc}$
$\Rightarrow\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3+\Big(\frac{-5}{6}\Big)^3=3\times\Big(\frac{1}{2}\Big)\times\Big(\frac{1}{3}\Big)\times\Big(\frac{-5}{6}\Big)$
$=\frac{-5}{12}$
$\therefore\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=\frac{-5}{12}$

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