Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{3z}+\frac{y^3}{3xz}+\frac{z^2}{3x}$?
Option 1: $0$
Option 2: $xz$
Option 3: $y$
Option 4: $3y$
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Correct Answer: $y$
Solution :
Given:
$x+y+z=0$
We know, $x^3+y^3+z^3-3xyz=0$
⇒ $x^3+y^3+z^3 = 3xyz$
Now, $\frac{x^2}{3z}+\frac{y^3}{3xz}+\frac{z^2}{3x}$
= $\frac{x^3+y^3+z^3 }{3xz}$
= $\frac{3xyz}{3xz}$
= $y$
Hence, the correct answer is $y$.
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