Question : If $x= a-b, y=b-c, z=c-a$, then the numerical value of the algebraic expression $x^3+y^3+z^3-3xyz$ will be:
Option 1: $a + b + c$
Option 2: $0$
Option 3: $4(a + b + c)$
Option 4: $3abc$
Correct Answer: $0$
Solution : Given: $x= (a-b), y=(b-c), z=(c-a)$ $⇒x+y+z=a-b+b-c+c-a=0$ We know that, if $ x+y+z=0$, then $x^3+y^3+z^3 -3xyz =0$ Because, $x^3+y^3+z^3 -3xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)$ Hence, the correct answer is $0$.
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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:
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:
Question : What will be the value of $x^{3}+y^{3}+z^{3}-3xyz$, when $x+y+z=9$ and $x^{2}+y^{2}+z^{2}=31?$
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:
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