Question 12 Marks
By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : $x^4 + 1; x - 1$
AnswerBy acute division, we have
quotient $= x^3 + x^2 + x + 1$ remainder $= 2$ View full question & answer→Question 22 Marks
Expand the following: $\Big(4-\frac{1}{3{\text{x}}}\Big)^3$
Answer$\Big(4-\frac{1}{3{\text{x}}}\Big)^3=(4)^3+\Big(-\frac{1}{3\text{x}}\Big)^3+ 3(4)\Big(-\frac{1}{3\text{x}}\Big)\Big(4-\frac{1}{3{\text{x}}}\Big)$
$[$Using identity, $(a - b)^3 = a^3 - b^3 + 3a(-b)(a - b)]$
$= 64 - \frac{1}{27\text{x}^3}-\frac{4}{\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$= 64 - \frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\frac{4}{3\text{x}^2}$
View full question & answer→Question 32 Marks
Factorise the following: $8\text{p}^3+\frac{12}{5}\text{p}^2+\frac{6}{25}\text{p}+\frac{1}{125}$
Answer$8\text{p}^3+\frac{12}{5}\text{p}^2+\frac{6}{25}\text{p}+\frac{1}{125}$
$=(2\text{p})^3+3\times(2\text{p})^2\times\frac{1}{5}+3\times(2\text{p})\times\Big(\frac{1}{5}\Big)^2+\Big(\frac{1}{5}\Big)^3$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$
$[$Using identity,$ (a - b)^3 = a^3 - b^3 + 3a(-b)(a - b)]$
$=\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$
View full question & answer→Question 42 Marks
Expand the following: $(3a - 2b)^3$
Answer$(3a - 2b)^3 = (3a)^3 + (-2b)^3 + 3(3a)(-2b)(3a - 2b)$
$[$Using identity, $(a - b)^3 = a^3 - b^3 + 3a(-b)(a - b)]$
$= 27a^3 - 8b^3 - 18ab(3a - 2b)$
$= 27a^3 - 8b^3 - 54a^2b + 36ab^2$
$= 27a^3 - 54a^2b + 36ab^2 - 8b^3$
View full question & answer→Question 52 Marks
Factorise the following: $16x^2 + 4y^2 + 9z^2 - 16xy - 12yz + 24xz$
Answer$16x^2 + 4y^2 + 9z^2 - 16xy - 12yz + 24xz$
$= (4x)^2 + (-2y)^2 + (3z)^2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x)$
$= {4x + (-2y) + 3z}^2 [\therefore a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2]$
$= (4x - 2y + 3z)^2$
$= (4x - 2y + 3z)(4x - 2y + 3z)$
View full question & answer→Question 62 Marks
Without finding the cubes, factorise: $(x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3$
AnswerWe know that,
$a^3 + b^3+ c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$
Also, if $a + b + c = 0,$ then $a^3 + b^3+ c^3 = 3abc$
Here, we see that$ (x - 2y) + (2y - 3z) + (3z - x) = 0$
Therefore,$ (x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3 $
$= 3(x - 2y)(2y - 3z)(3z - x) .$
View full question & answer→Question 72 Marks
If $\text{p(x)}= \text{x}^2-4\text{x}+3,$ evaluate:
$\text{p}(2)-\text{p}(-1)+\text{p}\big(\frac{1}{2}\big).$
AnswerWe have $\text{p(x)}= \text{x}^2-4\text{x}+3$
$\therefore$ $\text{p}(2)-\text{p}(-1)+\text{p}\Big(\frac{1}{2}\Big)$
$=(2^2-4\times2+3)-\big\{(-1)^2-4(-1)+3\big\}+\Big\{\Big(\frac{1}{2}\Big)^2-4\times\frac{1}{2}+3\Big\}$
$=(4-8+3)-(1+4+3)+\Big(\frac{1}{4}-2+3\Big)$
$=-1-8+\frac{5}{4}$
$=-9+\frac{5}{4}=\frac{-36+5}{4}=\frac{-31}{4}$
View full question & answer→Question 82 Marks
Check whether $p(x)$ is a multiple of $g(x) $or not: $p(x) = x^3 - 5x^2 + 4x - 3, g(x) = x - 2$
Answer$p(x)$ will be a multiple $g(x)$ if $g(x)$ divides $p(x).$
Now,$ g(x) = x - 2$ gives $x = 2$
Remainder $= p(2) = (2)^3 - 5(2)^2 + 4(2) - 3$
$= 8 - 5(4) + 8 - 3 = 8 - 20 + 8 - 3$
$= -7$
Since remainder $\neq 0,$ So $p(x)$ is not a multiple of $g(x)$.
View full question & answer→Question 92 Marks
Without actually calculating the cubes, find the value of: $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3$
AnswerGiven, $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3$ or $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3+\Big(-\frac{5}{6}\Big)^3$ Here, we see that, $\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$
$=\frac{3+2-5}{6}$
$=\frac{5-5}{6}=0$
$\therefore \ \Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3$
$=3\times\frac{1}{2}\times\frac{1}{3}\times\Big(-\frac{5}{6}\Big)$
$=-\frac{5}{12}$
$[$Using identity, if $a + b + c = 0,$ then $a^3 + b^3+ c^3 = 3abc]$
View full question & answer→Question 102 Marks
Find the following product: $(x^2 - 1) (x^4 + x^2 + 1)$
Answer$(x^2 - 1)(x^4 + x^2 + 1) $
$= (x^2 - 1) {(x^2)^2 + (x^2)(1) + (1)^2} $
$= (x^2)^3 - (1)^3 \big[\therefore (a + b)(a^2 - ab + b^2) = a^3 + b^3\big]$
$= x^6 - 1$
View full question & answer→Question 112 Marks
Using suitable identity, evaluate the following : $101 \times 102$
Answer$101 \times 102 = (100 + 1)(100 + 2)$ Now using identity $(x + a)(x + b) = x^2 + (a + b)x + ab,$ we have
$(100 + 1)(100 + 2) = (100)^2 + (1 + 2)100 + (1)(2)$
$= 10000 + (3)100 + 2 = 10000 + 300 + 2$
$= 10302$
View full question & answer→Question 122 Marks
Expand the following$: (3a - 5b - c)^2$
Answer$(3a - 5b - c)^2 = (3a)^2 + (-5b)^2 + (-c)^2 + 2(3a)^2 - 5b + 2(-5b)(-c) + 2(-c)(3a)$
[$\therefore a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2]$
$= 9a^2 + 25b^2 + c^2 - 30ab + 10bc + 6ca.$
View full question & answer→Question 132 Marks
Expand the following: $(4a - b + 2c)^2$
Answer$(4a - b + 2c)^2 = (4a)^2 + (-b)^2 + (2c)^2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)$
[$\therefore a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2]$
$= 16a^2 + b^2 + 4c^2 - 8ab - 4ac + 16ca$
View full question & answer→Question 142 Marks
Find the value of: $x^3 - 8y^3 - 36xy - 216,$ when $x = 2y + 6$
AnswerHere, we see that,$ x - 2y - 6 = 0 \therefore x^3 + (-2y)^3+ (-6)^3 = 3x(-2y)(-6) [$Using identity, if $a + b + c = 0,$ then $a^3 + b^{3 }+ c^3 = 3abc]$
$\Rightarrow x^3 - 8y^3 - 216 = 36xy ...(i)$
Now$, x^3 - 8y^3 - 36xy - 216$
$= x^3 - 8y^3 - 216 - 36xy$
$= 36xy - 36xy = 0 [$From $Eq...(i)]$
View full question & answer→Question 152 Marks
By Remainder Theorem find the remainder, when p(x) is divided by g(x), where:
$\text{p(x)}=\text{x}^3-6\text{x}^2+2\text{x}-4,$ and $\text{g(x)}=1-\frac{3}{2}\text{x} $
AnswerGiven, $\text{p(x)}=\text{x}^3-6\text{x}^2+2\text{x}-4,$ and $\text{g(x)}=1-\frac{3}{2}\text{x} $ Here, zero of g(x) is $\frac{2}{3}.$ When we divide p(x) by g(x) using remainder theorem, we get the remainder $\text{p}\big(\frac{2}{3}\big)$ $\therefore =\frac{8}{27}-6\times\frac{4}{9}+2\times\frac{2}{3}-4=\frac{8}{27}-\frac{24}{9}+\frac{4}{3}-4$ $=\frac{8-72+36-108}{27}=\frac{-136}{27}$
Hence, remainder is $\frac{-136}{27}.$
View full question & answer→Question 162 Marks
Factorise: $a^3 - 8b^3 - 64c^3 - 24abc$
AnswerWe have,
$a^3 - 8b^3 - 64c^3 - 24abc$
$= {(a)^3 + (-2b)^3 + (-4c)^3 - 3(a)(-2b)(-4c)}$
$= {a + (-2b) + (-4c)}{a^2 + (-2b)^2 + (-4c)^2 - a(-2b) - (-2b)(-4c) - (-4c)a}$
$\big[\therefore a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\big]$
$= (a + -2b + -4c)(a^2 + 4b^2 + 16c^2 + 2ab + 8bc + 4ca)$
View full question & answer→Question 172 Marks
Factorise the following: $25x^2 + 16y^2 + 4z^2 - 40xy + 16yz - 20xz$
Answer$25x^2 + 16y^2 + 4z^2 - 40xy + 16yz - 20xz $
$= (-5x)^2 + (4y)^2 + (2z)^2 + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x)$
$= (-5x + 4y + 2z)^2$
View full question & answer→Question 182 Marks
Factorise the following: $\Big(2\text{x}+\frac{1}{3}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2$
Answer$\Big(2\text{x}+\frac{1}{3}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2$
$=\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)-\Big(\text{x}-\frac{1}{2}\Big)\Bigg]\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)+\Big(\text{x}-\frac{1}{2}\Big)\Bigg]$
$[$Using identity, $a^2 - b^2 = (a - b)(a + b)]$
$=\Big(2\text{x}-\text{x}+\frac{1}{3}+\frac{1}{2}\Big)\Big(2\text{x}+\text{x}+\frac{1}{3}-\frac{1}{2}\Big)$
$=\Big(\text{x}+\frac{2+3}{6}\Big)\Big(3\text{x}+\frac{2-3}{6}\Big)$
$=\Big(\text{x}+\frac{5}{6}\Big)\Big(3\text{x}-\frac{1}{6}\Big)$
View full question & answer→Question 192 Marks
Find the following product: $\text{a}^3 - 2\sqrt{2}\text{b}^3$
Answer$\text{a}^3 - 2\sqrt{2}\text{b}^3=(\text{a})^3 - (\sqrt{2}\text{b})^3$
$(\text{a} - \sqrt{2}\text{b})\big\{(\text{a})^2 +(\text{a}) (\sqrt{2}\text{b})+(\sqrt{2}\text{b})^2\big\}$
$\big[\therefore a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$(\text{a} - \sqrt{2}\text{b})(\text{a}^2 + \sqrt{2}\text{ab}+2\text{b}^2)$
View full question & answer→Question 202 Marks
Factorise the following: $9x^2 + 4y^2 + 16z^2 + 12xy - 16yz - 24xz$
Answer$9x^2 + 4y^2 + 16z^2 + 12xy - 16yz - 24xz$
$= (3x)^2 + (2y)^2 + (-4z)^2 + 2(3x)(2y) + 2(y)(-4z) + 2(-4z)(3x)$
$= {3x + 2y (-4z)}^2 [\because a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2]$
$= (3x + 2y - 4z)2 = (3x + 2y - 4z)(3x + 2y - 4z)$
View full question & answer→Question 212 Marks
Factorise the following: $9x^2 - 12x + 4$
Answer$9x^2 - 12x + 4 = (3x)^2 - 2(3x)(2) + (2)^2$
$= (3x - 2)^2 [\therefore a^2 - 2ab + b^2 = (a - b)^2]$
$= (3x - 2)(3x - 2)$
View full question & answer→Question 222 Marks
Factorise:
$2\sqrt{2}\text{a}^3 + 8\text{b}^3 - 27\text{c}^3 + 18\sqrt{2}\text{abc}.$
AnswerWe have,
$2\sqrt{2}\text{a}^3 + 8\text{b}^3 - 27\text{c}^3 + 18\sqrt{2}\text{abc}.$
$=\big\{(\sqrt{2}\text{a})^3 + (2\text{b})^3 + (-3\text{c})^3 - 3(\sqrt{2}\text{a})(2\text{b})(-3\text{c})\big\}$
$=\big\{\sqrt{2}\text{a} + 2\text{b} -3\text{c}\big\}\big\{(\sqrt{2}\text{a})^2+(2\text{b})^2+(-3\text{c})^2-(\sqrt{2}\text{a})(2\text{b})-(2\text{b})(-3\text{c})(\sqrt{2}\text{a})\big\}$
$=(\sqrt{2}\text{a} + 2\text{b} -3\text{c})(2\text{a}^2+4\text{b}^2+9\text{c}^2-2\sqrt{2}\text{ab}+6\text{bc}+3\sqrt{2}\text{ca})$
View full question & answer→Question 232 Marks
For what value of m is $x^3 - 2\ mx^2 + 16$ divisible by $x + 2$ ?
AnswerLet $p(x) = x^3 - 2\ mx^2 + 16$
Since$, p(x)$ is divisible by $(x + 2),$ then remainder $= 0$
$P(-2) = 0$
$\Rightarrow (-2)^3 - 2m(-2)^2 + 16 = 0$
$\Rightarrow -8 - 8\ m + 16 = 0$
$\Rightarrow 8 = 8\ m$
$m = 1$
Hence, the value of $m$ is $1 $.
View full question & answer→Question 242 Marks
Give possible expressions for the length and breadth of the rectangle whose area is given by $4a^2 + 4a - 3.$
AnswerGiven, area of rectangle $= 4a^2 + 6a - 2a - 3$
$= 4a^2 + 4a - 3 [$by splitting middle term$]$
$= 2a(2a + 3) - 1(2a + 3) = (2a - 1)(2a + 3)$
Hence, possible length $= 2a - 1$ and breadth $= 2a + 3$
View full question & answer→Question 252 Marks
Find the following product: $\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$
Answer$\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$
$=\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big\{\Big(\frac{\text{x}}{2}\Big)^2-\Big(\frac{\text{x}}{2}\Big)(\text{2y})+(2\text{y})^2\Big\}$
$=\Big(\frac{\text{x}}{2}\Big)^3+(2\text{y})^3$ $\big[\therefore (a + b)(a^2 - ab + b^2) = a^3 + b^3\big]$
$=\frac{\text{x}}{8}^3+8\text{y}^3$
View full question & answer→Question 262 Marks
Show that: $x + 3$ is a factor of $69 + 11x - x^2 + x^3.$
AnswerLet $p(x) = 69 + 11x - x^2 + x^3, g(x) = x + 3.$
$g(x) = x + 3 = 0$ gives $x = -3$
$g(x)$ will be a factor of $p(x)$ if $p(-3) = 0($Factor theorem$)$
Now$, p(-3) = 69 + 11(-3) - (-3)^2 + (-3)^3$
$= 69 - 33 - 9 -27$
$= 0$
Since, $p(-3) = 0,$ So $g(x)$ is a factor of $p(x).$
View full question & answer→Question 272 Marks
Factorise: $1 + 64x^3$
AnswerWe have,
$1 + 64x^3 = (1)^3 + (4x)^3$
$= (1 + 4x){(1)^2 - (1)(4x) + (4x)^2}$
$\big[\therefore a^3 + b^3 = (a + b)(a^2 - ab + b^2)\big]$
$= (1 + 4x)(1 - 4x + 16x^2)$
$= (1 + 4x)(16x^2 -4x + 1)$
$= (4x + 1)(16x^2 -4x + 1)$
View full question & answer→Question 282 Marks
Find the zeroes of the polynomial: $p(x) = (x - 2)^2 - (x + 2)^2$
AnswerGiven, polynomial is $p(x) = (x - 2)^2 - (x + 2)^2$
For zeroes of polynomial, put$ p(x) = 0$
$(x – 2)^2 – (x+ 2)^2 = 0$
$(x - 2 + x + 2)(x - 2 - x - 2) = 0 [$using identity$, a^{2 }- b^2 = (a - b)(a + b)] $
$\Rightarrow (2x)(-4) = 0$
View full question & answer→Question 292 Marks
Factorise the following: $9x^2 - 12x + 3$
Answer$9x^2 - 12x + 3 = 9x^2 - 9x - 3x + 3$
$= 9x(x - 1) - 3(x - 1)$
$= (9x - 3)(x - 1)$
$= 3(3x - 1)(x - 1)$
View full question & answer→Question 302 Marks
By Remainder Theorem find the remainder, when $p(x)$ is divided by $g(x),$ where:$ p(x) = 4x^3 - 12x^2 + 14x - 3, g(x) = 2x - 1$
AnswerGiven, $p(x) = 4x^3 - 12x^2 + 14x - 3, g(x) = 2x - 1$
Here, zero of $g(x)$ is $\frac{1}{2}.$
When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $\text{p}\big(\frac{1}{2}\big)$
$\therefore\text{p}\big(\frac{1}{2}\big)=4\big(\frac{1}{2}\big)^3-12\big(\frac{1}{2}\big)^2+14\big(\frac{1}{2}\big)-3$
$=4\times\frac{1}{8}-12\times\frac{1}{4}+14\times\frac{1}{2}-3$
Hence, remainder is $\frac{3}{2}.$
View full question & answer→Question 312 Marks
Find the value of: $x^3 + y^3 - 12xy + 64,$ when $x + y = -4$
AnswerHere, we see that, $x + y + 4 = 0$
$\therefore x^3 + y^3+ (4)^3 = 3xy(4)$
$[$Using identity, if $a + b + c = 0,$ then$ a^3 + b^{3 }+ c^3 = 3abc]$
$= 12xy ...(i)$
Now$, x^3 + y^3 - 12xy + 64 = x^3 + y^3+ 64 - 12xy$
$= 12xy - 12xy = 0 [$From $Eq...(i)]$
View full question & answer→Question 322 Marks
By Remainder Theorem find the remainder, when $p(x)$ is divided by $g(x),$ where: $p(x) = x^3 - 2x^2 - 4x - 1, g(x) = x + 1$
AnswerGiven, $p(x) = x^3 - 2x^2 - 4x - 1$ and $g(x) = x + 1$
Here, zero of $g(x)$ is $-1$.
When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p(-1)$
$\therefore p(-1) = (-1)^3 - 2(-1)^2 - 4(-1) - 1$
$= -1 - 2 + 4 - 1$
$= 4 - 4 = 0$
Hence, remainder is $0.$
View full question & answer→Question 332 Marks
Without actually calculating the cubes, find the value of: $(0.2)^3 - (0.3)^3 + (0.1)^3$
AnswerGiven,$ (0.2)^3 - (0.3)^3 + (0.1)^3 or (0.2)^3 + (-0.3)^3 + (0.1)^3$
Here, we see that, $0.2 - 0.3 + 0.1 = 0.3 - 0.3 = 0$
$\therefore (0.2)^3 - (0.3)^3 + (0.1)^3 = 3 \times (0.2) \times (-0.3) \times (0.1)$
$[$Using identity, if $a + b + c = 0,$ then $a^3 + b^3+ c^3 = 3abc]$
$= -0.6 \times 0.003 = -0.018$
View full question & answer→Question 342 Marks
By Remainder Theorem find the remainder, when $p(x)$ is divided by $g(x),$ where: $p(x) = x^3 - 3x^2 + 4x + 50, g(x) = x - 3$
AnswerGiven, $p(x) = x^3 - 3x^2 + 4x + 50, g(x) = x - 3$
Here, zero of $g(x)$ is $3$.
When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p(3)$
$\therefore p(3) = (3)^3 - 3(3)^2 + 4(3) + 50$
$= 27 - 27 + 12 + 50 = 62$
Hence, remainder is $62$.
View full question & answer→Question 352 Marks
Using suitable identity, evaluate the following: $103^3$
Answer$103^3 = (100 + 3)^3$
Now using identity$ (a + b)^3 = a^3 + b^3 + 3ab(a + b),$ we have
$(100 + 3)^3 = (100)^3 + (3)^3 + 3(100)(3)(100 + 3)$
$= 1000000 + 27 + 900(100 + 3)$
$= 1000000 + 27 + 90000 + 2700$
$= 1092727$
View full question & answer→Question 362 Marks
Expand the following: $(-x + 2y - 3z)^2$
Answer$(-x + 2y - 3z)^2 = {(-x) + 2y + (-3z)}^2$
$= (-x)^2 + (2y)^{2 }+ (-3z)^{2 }+ 2(-x)(2y) + 2(2y)(-3z) + 2(-3yz)(-x)$
$= x^2 + 4y^2 + 9z^2 - 4xy - 12yz + 6xz.$
View full question & answer→Question 372 Marks
Expand the following: $\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$
Answer$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}}\Big)^3 + \Big(\frac{\text{y}}{3}\Big)^3 + 3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$[$Using identity,$ (a - b)^3 = a^3 - b^3 + 3a(-b)(a - b)]$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}}=\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}$
View full question & answer→Question 382 Marks
If $a, b, c$ are all non$-$zero and $a + b + c = 0,$ prove that $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3.$
AnswerWe have $a, b, c$ are all non-zero and $a + b + c = 0,$ therefore
$a^3 + b^{3 }+ c^3 = 3abc$
Now, $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}=\frac{3\text{abc}}{\text{abc}}=3$
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