Question

find the value of a^3+b^3+6ab-8,when a+b=2

Answer 1

$a + b = 2 \\ cube \: on \: both \: sides \: \\ (a + b) ^{3} = 2 ^{3} \\ a ^{3} + b ^{3} + 3ab(a + b) = 8 \\ a ^{3} + b ^{3} + 3ab(2) = 8 \\ a ^{3} + b ^{3} + 6ab - 8 = 0$