Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$?
Option 1: $0$
Option 2: $\frac{1}{3}$
Option 3: $1$
Option 4: $3$
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Correct Answer: $3$
Solution : Given: $x+y+z=0$ So, $x^3+y^3+z^3=3xyz$ $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$ = $\frac{x^3}{xyz}+\frac{y^3}{xyz}+\frac{z^3}{xyz}$ = $\frac{x^3+y^3+z^3}{xyz}$ = $\frac{3xyz}{xyz}$ = $3$ Hence, the correct answer is $3$.
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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:
Option 1: 3
Option 2: 1
Option 3: 0
Option 4: 2
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 3: 2
Option 4: 3
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
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 2: $xz$
Option 3: $y$
Option 4: $3y$
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Option 2: $1$
Option 3: $2$
Option 4: $–1$
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