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 $x=5$ ; $y=6$ and $z=-11$, then the value of $x^{3}+y^{3}+z^{3}$ is:

Option 1:  –890

Option 2:  –970

Option 3:  –870

Option 4:  –990

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 $xy+yz+zx=0$, 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 : If $x+y+z=13$ and $x^2+y^2+z^2=69$, then $xy+z(x+y)$ is equal to:

Option 1: 70

Option 2: 40

Option 3: 50

Option 4: 60