Find the value of $a^3+b^3+c^3-3abc$, when $a+b+c=5, a^2+b^2+c^2=11$

Show that ${\left(1-a\right)}^3+{\left(b-1\right)}^3+{\left(a-b\right)}^3=3\left(1-a\right)\left(b-1\right)\left(a-b\right)$ 

If $x=2-k,$prove that $x^3+6xk+k^3-8=0$

If $x+y+z=0$, then prove that $x^3+y^3+z^3=3xyz$

Prove that $\frac{{\left(a^2-b^2\right)}^3+{\left(b^2-c^2\right)}^3+{(c^2-a^2)}^3}{{\left(a-b\right)}^3+{\left(b-c\right)}^3+{\left(c-a\right)}^3}=(a+b)(b+c)(c+a)$

Factorise:

$x^2-x-20$

Factorise:

$x^2-12x+36$

Factorise:

$x^2-20x+91$

Factorise:

$x^2-x-420$