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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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 2: $\frac{1}{3}$
Option 3: $1$
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
Option 3: 2
Option 4: 3
Question : If $x+y+z=0$, then what is the value of $\frac{\left (3y^{2}+x^{2}+z^{2} \right )}{\left (2y^{2}-xz \right)}$?
Option 1: $2$
Option 2: $1$
Option 3: $\frac{3}{2}$
Option 4: $\frac{5}{3}$
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Option 3: $2$
Option 4: $–1$
Question : If $\frac{1}{x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}}}=\frac{29}{79}$, where x, y, and z are natural numbers, then the value of $(2 x+3 y-z)$ is:
Option 1: 0
Option 2: 4
Option 3: 1
Option 4: 2
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