Question : If $x^2 = y+z$, $y^2=z+x$, $z^2=x+y$, then the value of $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$ is:

Option 1: –1

Option 2: 1

Option 3: 2

Option 4: 4

Question : If $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$, then what is the value of $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$?

Option 1: $0$

Option 2: $1$

Option 3: $\frac{3}{2}$

Option 4: $3$

Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}$, where $x \neq y \neq z$, is:

Option 1: $0$

Option 2: $\frac{1}{(x+y+z)}$

Option 3: $\frac{1}{(x+y)(y+z)(z+x)}$

Option 4: $1$

Question : What is $\frac{\left (x^{2}-y^{2} \right)^{3}+\left (y^{2}-z^{2} \right )^{3}+\left (z^{2}-x^{2} \right )^{3}}{\left (x-y \right)^{3}+\left (y-z \right )^{3}+\left (z-x \right)^{3}}?$

Option 1: $\frac{(x+y)(y+z)}{(x+z)}$

Option 2: $(x+y)^3(y+z)^3(z+x)^3$

Option 3: $(x+y)(y+z)(z+x)$

Option 4: $(x+y)(y+z)$

Question : If $x+y+z=17, x y z=171$ and $x y+y z+z x=111$, then the value of $\sqrt[3]{\left(x^3+y^3+z^3+x y z\right)}$ is:

Option 1: –64

Option 2: 4

Option 3: 0

Option 4: –4