Question

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

Answer 1

Heya mate
here is your answer

I hope this will help you...✌️✌️
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Answer 2

$\textbf{Your answer is --}$

Given, $\textbf{p = (2-a) }$

Let,

LHS = a^3 + 6ap + p^3 - 8

= (a^3 + p^3) + 6ap - 8

= (a+p) ( a^2+p^2-ap) + 6ap - 8

Now, put the value of p , we get

= (a+2-a) {a^2+(2-a)^2 - a(2-a)} +6a(2-a)-8

= 2 (a^2+4+a^2-4a - 2a +a^2)+12a-6a^2-8

= 2(3a^2 -6a+4) + 12a - 6a^2 - 8

= 6a^2 - 12a + 8 + 12a - 6a^2 - 8

= 0 = RHS

$\boxed{Hence\:proved }$

$\boxed{Hope\:it\:helps\:you!}$