Question 13 Marks
Expand : $(6a – 7b)^3$
Answer$(6a – 7b)^3= (6a)^3 – (7b)^3 – 3(6a) (7b) (6a – 7b)$
$= 216a^3 – 343b^3– 126ab (6a – 7b)$
$= 216a^3 – 343b^3 – 756a^2b + 882ab^2$
$= 216a^3 – 756a^2b + 882ab^2 – 343b^3$
View full question & answer→Question 23 Marks
If $2 \mathrm{a}+\frac{1}{2 \mathrm{a}}=8$, find $: 4 \mathrm{a}^2+\frac{1}{4 \mathrm{a}^2}$
Answer$ 4 a^2+\frac{1}{4 a^2}=\left(2 a+\frac{1}{2 a}\right)^2-2.2 a \frac{1}{2 a}$
$ \left(2 a+\frac{1}{2 a}\right)^2-2$
$ =(8)^2-2$
$ =64-2$
$ =62$
View full question & answer→Question 33 Marks
If $a^2 + b^2 = 41$ and $ab = 4,$ find : $a + b$
Answer$(a + b)^2 = a^2 + b^2 + 2ab= 41 + 2(4)$
$= 41 + 8$
$= 49$
$\Rightarrow (a + b)^2 = 49$
$\therefore a + b = 7$
View full question & answer→Question 43 Marks
Find the square of $607$
Answer$(607)^2 = (600 + 7)^2= (600)^2 + (7)^2 + 2 (600) (7)$
$= 360000 + 49 + 8900$
$= 368449$
View full question & answer→Question 53 Marks
Evaluate: $203 \times 197$
Answer$203 \times 197= (200 + 3) (200 − 3)$
$= (200)^2− (3)^2 ......[ \because (a − b) (a + b) = a^2 − b^2]$
$= 40000 − 9$
$= 39991$
View full question & answer→Question 63 Marks
Evaluate: $20.8 \times 19.2$
Answer$20.8 \times 19.2= (20 + 0.8) (20 − 0.8)$
$= (20)^2− (0.8)^2 ......[ \because (a − b) (a + b) = a^2 − b^2]$
$= 400 − 0.64$
$= 399.36$
View full question & answer→Question 73 Marks
Evaluate: $(a + bc) (a − bc) (a^2 + b^2c^2)$
Answer($a + bc) (a − bc) (a^2 + b^2c^2)= [a^2 − (bc)^2] (a^2 + b^2c^2) .......[ \because (a − b) (a + b) = a^2 − b^2]$
$= (a^2 − b^2c^2) (a^2+ b^2c^2)$
$= (a^2)^2 − (b^2c^2)^2 .......[ \because (a − b) (a + b) = a^2 − b^2]$
$= a^4 − b^4c^4$
View full question & answer→Question 83 Marks
Evaluate: $(3x + 4y) (3x − 4y) (9x^2 + 16y^2)$
Answer$(3x + 4y) (3x − 4y) (9x^2 + 16y^2)= [(3x)^2 − (4y)^2] (9x^2 + 16y^2) .....[ \because (a − b) (a + b) = a^2 − b^2]$
$= (9x^2 − 16y^2) (9x^2 + 16y^2)$
$= (9x^2)^2 − (16y^2)^2 ......[ \because (a − b) (a + b) = a^2− b^2]$
$= 81x^4 − 256y^4$
View full question & answer→Question 93 Marks
Evaluate: $(m + 3) (m − 3) (m^2 + 9)$
Answer$(m + 3) (m − 3) (m^2 + 9)= (m)^2− (3)^2 (m^2 + 9) ......[ \because (a − b) (a + b) = a^2 − b^2]$
$= (m^2− 9) (m^2 + 9)$
$= (m^2)^2− 9^2$
$= m^4− 81$
View full question & answer→Question 103 Marks
Evaluate: $(3a^2 − 4b^2) (8a^2 − 3b^2)$
Answer$(3a^2 − 4b^2) (8a^2 − 3b^2)= 3a^2(8a^2 − 3b^2) − 4b^2 (8a^2 − 3b^2)$
$= 24a^4 − 9a^2b^2 − 32a^2b^2 + 12b^4$
$= 24a^4− 41a^2b^2 + 12b^4$
View full question & answer→Question 113 Marks
If $5x – 4y = 7$ and $xy = 8$, find : $125x^3 – 64y^3.$
Answer$125x^3 − 64y^3 = (5x)^3 − (4y)^3= (5x − 4y)^3 + 3(5x) (4y) (5x − 4y)$
$= (5x − 4y)^3 + 60xy (5x − 4y)$
$= (7)^3+ 60 (8) (7)$
$= 343 + 3360$
$= 3703$
View full question & answer→Question 123 Marks
If $3x + 2y = 9$ and $xy = 3,$ find : $27x^3 + 8y^3.$
Answer$27x^3 + 8y^3 = (3x)^3 + (2y)^3= (3x + 2y)^3 − 3.3x . 2y (3x + 2y)$
$= (3x − 2y)^3 − 18xy (3x + 2y)$
$= (9)^3− 18(3) (9)$
$= 729 − 486$
$= 243$
View full question & answer→Question 133 Marks
If $3x – 4y = 5$ and $xy = 3$, find : $27x^3 – 64y^3.$
Answer$27x^3 – 64x^3 = (3x)^3 – (4y)^3= (3x − 4y)^3 (3x − 4y)^3 + 3 (3x) (4y) (3x − 4y) ......[ \because a^3 − b^3= (a − b)^3 + 3ab(a − b)]$
$= (5)^3 + 36(xy) (3x − 4y)$
$= 125 + 36 (3) (5)$
$= 125 + 540$
$= 665$
View full question & answer→Question 143 Marks
If $a + b + c = 11$ and $a^2 + b^2 + c^2 = 81$, find : $ab + bc + ca.$
AnswerSince $(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$
\therefore(11)^2=81+2(a b+b c+c a)
$
$
\therefore 2(a b+b c+c a)=121-81=40
$
$a b+b c+c a=\frac{40}{2}$
$
\Rightarrow \mathrm{ab}+\mathrm{bc}+\mathrm{ca}=20
$
View full question & answer→Question 153 Marks
Evaluate: $\left(3 \mathrm{x}+\frac{1}{2}\right)\left(2 \mathrm{x}+\frac{1}{3}\right)$
Answer$ \left(3 \mathrm{x}+\frac{1}{2}\right)\left(2 \mathrm{x}+\frac{1}{3}\right)$
$ =3 \mathrm{x}\left(2 \mathrm{x}+\frac{1}{3}\right)+\frac{1}{2}\left(2 \mathrm{x}+\frac{1}{3}\right)$
$ =6 \mathrm{x}^2+\mathrm{x}+\mathrm{x}+\frac{1}{6}$
$ =6 \mathrm{x}^2+2 \mathrm{x}+\frac{1}{6}$
View full question & answer→Question 163 Marks
If $a + b + c = 9$ and $ab + bc + ca = 15$, find: $a^2 + b^2 + c^2.$
AnswerSince $(a + b + c)^2= a^2 + b^2 + c^2 + 2 (ab + bc + ca)\therefore (9)^2 = a^2 + b^2+ c^2 + 2 (15)$
$81 = a^2+ b^2 + c^2 + 30$
$\therefore a^2 + b^2 + c^2= 81 − 30 = 51$
View full question & answer→Question 173 Marks
If $\mathrm{a}^2+\frac{1}{\mathrm{a}^2}=11$, find : $\mathrm{a}-\frac{1}{\mathrm{a}}$
Answer$ \left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}-2$
$ \Rightarrow\left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=11-2$
$ \Rightarrow\left(\mathrm{a}-\frac{1}{\mathrm{a}}\right)^2=9$
$ \Rightarrow \mathrm{a}-\frac{1}{\mathrm{a}}=\sqrt{9}$
$ \Rightarrow \mathrm{a}-\frac{1}{\mathrm{a}}= \pm 3$
View full question & answer→Question 183 Marks
If $\mathrm{a}^2+\frac{1}{\mathrm{a}^2}=23$, find $: \mathrm{a}+\frac{1}{\mathrm{a}}$
Answer$ \left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^2=\mathrm{a}^2+\frac{1}{\mathrm{a}^2}+2$
$ \Rightarrow\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^2=23+2$
$ \Rightarrow\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^2=25$
$ \Rightarrow \mathrm{a}+\frac{1}{\mathrm{a}}=\sqrt{25}$
$ \Rightarrow \mathrm{a}+\frac{1}{\mathrm{a}}= \pm 5$
View full question & answer→Question 193 Marks
If $a^2 + b^2= 10$ and $ab = 3$; find : $a + b$
Answer$ (a+b)^2=a^2+b^2+2 a b$
$ \Rightarrow(a+b)^2=10+2 \times 3$
$ \Rightarrow(a+b)^2=10+6$
$ \Rightarrow(a+b)^2=16$
$ \Rightarrow(a+b)=\sqrt{16}$
$ \Rightarrow a+b= \pm 4$
View full question & answer→Question 203 Marks
If $a^2 + b^2= 10$ and $ab = 3;$ find : $a – b$
Answer$ (a-b)^2=a^2+b^2-2 a b$
$ \Rightarrow(a-b)^2=10-2 \times 3$
$ \Rightarrow(a-b)^2=10-6$
$ \Rightarrow(a-b)^2=4$
$ \Rightarrow(a-b)=\sqrt{4}$
$ \Rightarrow a-b= \pm 2$
View full question & answer→Question 213 Marks
If $a^2 + b^2 = 29$ and $ab = 10,$ find : $a − b$
Answer$ (a-b)^2=a^2+b^2-2 a b$
$ \Rightarrow(a-b)^2=29-2 \times 10$
$ \Rightarrow(a-b)^2=29-20$
$ \Rightarrow(a-b)^2=9$
$ \Rightarrow a-b=\sqrt{9}$
$ \Rightarrow a-b= \pm 3$
View full question & answer→Question 223 Marks
If $a^2 + b^2 = 29$ and $ab = 10,$ find : $a + b$
Answer$(a+b)^2=a^2+b^2+2 a b$
$\Rightarrow(a+b)^2=29+2 \times 10$
$\Rightarrow(a+b)^2=29+20$
$\Rightarrow(a+b)^2=49$
$\Rightarrow a+b=\sqrt{49}$
$\Rightarrow \mathrm{a}+\mathrm{b}= \pm 7$
View full question & answer→Question 233 Marks
If $a – b = 6$ and $ab = 16$, find $a^2 + b^2$
Answer$(a − b)^2 = a^2+ b^2 − 2ab$
$\Rightarrow (6)^2= a^2 + b^2 − 2\times 16$
$\Rightarrow 36 = a^2 + b^2 − 32$
$\Rightarrow 36 + 32 = a^2 + b^2$
$\Rightarrow 68 = a^2 + b^2$
$\therefore a^2 + b^2 = 68$
View full question & answer→Question 243 Marks
If $a+b=5$ and $ab = 6,$ find $a^2 + b^2$
Answer$(a + b)^2 = a^2+ b^2 + 2ab$
$\Rightarrow (5)^2 = a^2 + b^2+ 2\times 6$
$\Rightarrow 25 = a^2 + b^2 + 12$
$\Rightarrow 25 − 12 = a^2 + b^2$
$\Rightarrow 13 = a^2 + b^2$
$\therefore a^2 + b^2 = 13$
View full question & answer→Question 253 Marks
Find the cube of: $x-\frac{1}{2}$
Answer$ \left(\mathrm{x}-\frac{1}{2}\right)^3$
$ =(\mathrm{x})^3-\left(\frac{1}{2}\right)^3-3 \times \mathrm{x} \times \frac{1}{2}\left(\mathrm{x}-\frac{1}{2}\right)$
$ =\mathrm{x}^3-\frac{1}{8}-\frac{3 \mathrm{x}}{2}\left(\mathrm{x}-\frac{1}{2}\right)$
$ =\mathrm{x}^3-\frac{1}{8}-\frac{3 \mathrm{x}^2}{2}+\frac{3 \mathrm{x}}{4}$
$ =\mathrm{x}^3-\frac{3 \mathrm{x}^2}{2}+\frac{3 \mathrm{x}}{4}-\frac{1}{8}$
View full question & answer→Question 263 Marks
Find the cube of: $2 x+\frac{1}{x}$
Answer$ \left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)^3$
$ =(2 \mathrm{x})^3+\left(\frac{1}{\mathrm{x}}\right)^3+3 \times 2 \mathrm{x} \times \frac{1}{\mathrm{x}}\left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)$
$ =8 \mathrm{x}^3+\frac{1}{\mathrm{x}^3}+6\left(2 \mathrm{x}+\frac{1}{\mathrm{x}}\right)$
$ =8 \mathrm{x}^3+\frac{1}{\mathrm{x}^3}+12 \mathrm{x}+\frac{6}{\mathrm{x}}$
$ =8 \mathrm{x}^3+12 \mathrm{x}+\frac{6}{\mathrm{x}}+\frac{1}{\mathrm{x}^3}$
View full question & answer→Question 273 Marks
Expand: $\left(a+\frac{1}{a}\right)^3$
Answer$ \left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)^3$
$ =\mathrm{a}^3+\left(\frac{1}{\mathrm{a}}\right)^3+3 \times \mathrm{a} \times \frac{1}{\mathrm{a}} \times\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)$
$ =\mathrm{a}^3+\frac{1}{\mathrm{a}^3}+3\left(\mathrm{a}+\frac{1}{\mathrm{a}}\right)$
$ =\mathrm{a}^3+\frac{1}{\mathrm{a}^3}+3 \mathrm{a}+\frac{3}{\mathrm{a}}$
View full question & answer→Question 283 Marks
Evaluate : $(2a +3) (2a − 3) (4a^2 + 9)$
Answer$(2a +3) (2a − 3) (4a^2 + 9)$
$= [(2a)^2 − (3)^2] (4a^2 + 9) ..........[(a+b) (a−b) = a^2 − b^2]$
$= (4a^2 − 9) (4a^2 + 9)$
$= (4a^2)^2 − (9)^2 .........[(a+b) (a−b) = a^2 − b^2]$
$= 16a^4 − 81$
View full question & answer→