Question : Simplify the given expression. $\frac{x^3+y^3+z^3-3 x y z}{(x-y)^2+(y-z)^2+(z-x)^2}$

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

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

Option 3: $\frac{1}{4}(x+y+z)$

Option 4: $\frac{1}{2}(x+y+z)$

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: $0$

Question : What is the value of $\frac{4x^2+9y^2+12xy}{144}$?

Option 1: $(\frac{x}{3} + \frac{y}{4})^2$

Option 2: $(\frac{x}{3} + y)^2$

Option 3: $(\frac{x}{4} + \frac{y}{6})^2$

Option 4: $(\frac{x}{6} + \frac{y}{4})^2$

Question : If ${x^2+y^2+z^2=2(x+z-1)}$, then the value of $x^3+y^3+z^3$ is equal to:

Option 1: 6

Option 2: 1

Option 3: 2

Option 4: 8

Question : If $x(x+y+z)=20$, $y(x+y+z)=30$, and $z(x+y+z)=50$, then the value of $2(x+y+z)$ is:

Option 1: 20

Option 2: –10

Option 3: 15

Option 4: 18