Question

find the value of a cube plus b cube + 6ab - 8, when a + b = 2

Answer 1

$(a + b) ^3 = {a}^{3} + {b}^{3} + 3ab(a + b) \\ \\ after \: put \: \\ the \: value \\ \\ {2}^{3} = {a}^{3} + {b}^{3} + 3ab(2) \\ \\ 8 = {a}^{3} + {b}^{3}+ 6ab \\ then \: now \\ {a}^{3} + {b}^{3} + 6ab - 8 \\ = 8 - 8 \\ = 0$