Question

If p = 2- a, prove that a3+6ap+p3- 8 = 0

Answer 1

Step-by-step explanation:

$\sf\ p = 2 - a \\ \\ \sf\ {a}^{3} + 6ap + {p}^{3} - 8 \\ \\ \sf\ \implies: {a}^{3} + 6a(2 - a) + {(2 - a)}^{3} - 8 \\ \\ \sf\ \implies: {a}^{3} + 12a { - 6a}^{2} + {2}^{ 3} - 3. {2}^{2}.a + 3.2. {a}^{2} - {a}^{3} - 8 \\ \\ \sf\ \implies: {a}^{3} + 12a { - 6a}^{2} + 8 {- 12a} + {6a}^{2} { - a}^{3} - 8 \\ \\\sf\ \implies: {a}^{3} { - a}^{3} + 12a - 12a { - 6a }^{2} + {6a}^{2} + 8 - 8 \\ \\ \huge{\sf\ \implies:0 } \: \: \: \ddot \smile \\ \\ \sf\ \: \: \: \: \underline{\red{prove}}$