Question

If p=2-a prove that a3+6ap+p3=0

Answer 1

The answer which come after solving out is =8

Answer 2

$if \: p = 2 - a \\ p + a = 2 \\ cubing \\ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \\ {(p + a)}^{3} = {p}^{3} + {a}^{3} + 3ap(a + p) \\ put \: value \: of \: p + a \\ {2 }^{3} = {p}^{3} + {a}^{3} + 3ap(2) \\ 8 = {p}^{3} + {a}^{3} + 6ap \\ i \: think \: its \: not \: 0 \: its \: 8$