Question : If $\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3$, then what is the value of $(x+y+z)^3$?
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3
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Correct Answer: 0
Solution : Given:$\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3$ We know the identity, if $(x+y+z)=0$, then $(x^3+y^3+z^3)=3xyz$ Take the LCM of the given expression, we get– $\frac{x^3+y^3+z^3}{xyz}=3$ ${x^3+y^3+z^3}=3(xyz)$ So, $(x+y+z)=0$ The value of $(x+y+z)^3=0$ Hence, the correct answer is 0.
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Question : If $x+y+z=0$, then the value of $\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}$ is:
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$?
Question : If $xy+yz+zx=1$ , then the value of $\frac{1\:+\:y^2}{(x\:+\:y)(y\:+\:z)}$ is:
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Question : If $x+y+z = 22$ and $xy+yz+zx = 35$, then what is the value of $\small (x-y)^{2}+(y-z)^{2}+(z-x)^{2}$?
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