If* x+y+z=0*, then \(\frac{x^2 + yz}{yz}+\frac{y^2 + zx}{zx}+\frac{z^2 + xy}{xy}=?\)

## Answer

Correct Answer : b ) 6

Explanation :Consider the given expression

\(\frac{x^2 +yz}{yz}+\frac{y^2 + zx}{zx}+\frac{z^2 + xy}{xy}\)

or

\(=\frac{x^3+xyz}{xyz}+\frac{y^3+xyz}{xyz}+\frac{z^3+xyz}{xyz}\)

\(=\frac{x^3+y^3+z^3+3xyz}{xyz}\)

As x+y+z=0

So, x^{3}+y^{3}+z^{3}= 3xyz

Thus, putting the value of (x^{3}+y^{3}+z^{3}) in our expression we get

\(\Rightarrow \frac{x^3+y^3+z^3+3xyz}{xyz}\)

\(\Rightarrow \frac{3xyz+3xyz}{xyz}\)

= 6

Hence, (b) is the correct answer.

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