2015/114) Solve in real $a^3 + b^3 = 8 – 6ab$

$a^3 + b^3 - 8 + 6ab = 0$

or

$a^3 + b^3 + (-2)^3 -3(-2)(a)(b) = 0$

=> $a + b – 2 = 0$ or $a=b= - 2$

using the fact

$x^3 + y^3 + z^3 – 3xyz = \dfrac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$