Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Option 1: 1
Option 2: 0
Option 3: 2
Option 4: 3
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Correct Answer: 3
Solution :
Given: $x+y+z=0$
Cubing both sides, we get,
$⇒(x+y+z)^3=0$
$⇒x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)=0$
$⇒x^3+y^3+z^3 = -3(x+y)(y+z)(z+x)$ ......(1)
From the given equation ($x+y+z=0$), we can get,
$x+y=-z,$ $y+z=-x$ and $z+x=-y$
Putting in (1), we get,
$x^3+y^3+z^3 = -3(-z)(-x)(-y)=3xyz$
Consider, $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$
$=\frac{x^2(xz)(xy)+y^2(yz)(xy)+z^2(yz)(xz)}{(yz)(xz)(xy)}$
$=\frac{x^4yz+y^4zx+z^4xy}{x^2y^2z^2}$
Taking $xyz$ as common, we get,
$=\frac{xyz(x^3+y^3+z^3)}{x^2y^2z^2}$
$=\frac{x^3+y^3+z^3}{xyz}$
$=\frac{3xyz}{xyz}$
$=3$
Hence, the correct answer is 3.
Related Questions
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$?
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