Question : If $xy+yz+zx=0$, then $(\frac{1}{x^2–yz}+\frac{1}{y^2–zx}+\frac{1}{z^2–xy})$$(x,y,z \neq 0)$ is equal to:
Option 1: $3$
Option 2: $1$
Option 3: $x+y+z$
Option 4: $0$
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Correct Answer: $0$
Solution :
Given: $xy+yz+zx=0$ -----------------------------(1)
So, $xy+yz=-zx, yz+zx=-xy, zx+xy=-yz$
Now, $\frac{1}{x^2–yz}+\frac{1}{y^2–zx}+\frac{1}{z^2–xy}$
= $\frac{1}{x^2+zx+xy}+\frac{1}{y^2+xy+yz}+\frac{1}{z^2+yz+zx}$ (Putting the values of $-yz,-zx$ and $-xy$)
= $\frac{1}{x(x+z+y)}+\frac{1}{y(y+x+z)}+\frac{1}{z(z+y+x)}$
= $\frac{yz+zx+xy}{xyz(x+y+z)}$
= $\frac{0}{xyz(x+y+z)}$ (From equation (1))
= 0
Hence, the correct answer is $0$.
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