Question : What will be the value of $x^{3}+y^{3}+z^{3}-3xyz$, when $x+y+z=9$ and $x^{2}+y^{2}+z^{2}=31?$
Option 1: 27
Option 2: 3
Option 3: 54
Option 4: 9
Correct Answer: 54
Solution : Give: $x+y+z=9$ ......(1) $x^{2}+y^{2}+z^{2}=31$ ......(2) As $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$ Squaring (1) on both sides, we have, $(x+y+z)^2 = 81$ ⇒ $x^2 + y^2 + z^2 +2(xy+yz+zx) = 81$ ⇒ $31+2(xy+yz+zx) = 81$ ⇒ $xy+yz+zx = 25$ Now, $x^{3}+y^{3}+z^{3}-3xyz$ = $ (x+y+z)(x^2 + y^2 + z^2 - xy - yz- zx)$ = $(9)(31-25) = 9\times 6 = 54$ Hence, the correct answer is 54.
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Question : If $x+y+z=13,x^2+y^2+z^2=133$ and $x^3+y^3+z^3=847$, then the value of $\sqrt[3]{x y z}$ is:
Option 1: $8$
Option 2: $7$
Option 3: $-9$
Option 4: $-6$
Question : If $x^2 = y+z$, $y^2=z+x$, $z^2=x+y$, then the value of $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$ is:
Option 1: –1
Option 2: 1
Option 3: 2
Option 4: 4
Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}$, where $x \neq y \neq z$, is:
Option 1: $0$
Option 2: $\frac{1}{(x+y+z)}$
Option 3: $\frac{1}{(x+y)(y+z)(z+x)}$
Option 4: $1$
Question : What is $\frac{\left (x^{2}-y^{2} \right)^{3}+\left (y^{2}-z^{2} \right )^{3}+\left (z^{2}-x^{2} \right )^{3}}{\left (x-y \right)^{3}+\left (y-z \right )^{3}+\left (z-x \right)^{3}}?$
Option 1: $\frac{(x+y)(y+z)}{(x+z)}$
Option 2: $(x+y)^3(y+z)^3(z+x)^3$
Option 3: $(x+y)(y+z)(z+x)$
Option 4: $(x+y)(y+z)$
Question : If $x+y+z=17, x y z=171$ and $x y+y z+z x=111$, then the value of $\sqrt[3]{\left(x^3+y^3+z^3+x y z\right)}$ is:
Option 1: –64
Option 2: 4
Option 3: 0
Option 4: –4
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