Question

hyy gyzz


solve this question.

if P = 2 - a, prove that a^3+ 6ap + p^3 -8 = 0​

Answer 1

Explanation:

Given :-

p=2-a

To Prove :-

${\mathrm{a^3+6ap+p^3-8=0\:\:\:\:\:\:\: ........... (i) }}$

Proof :-

${\large {\mathrm {p=2-a\: put \: in\: (i) \: equation}}}$

$\large {\mathrm{a^3+6a(2-a)+{(2-a)}^3-8} \\ {\implies a^3+12a-6a^2+(2-a)(2^2 + a^2 - 2 \times 2 \times a) - 8} \\ {\implies a^3+12a-6a^2+(2-a)(4+a^2-4a)-8} \\ {\implies a^3+12a-6a^2+8+2a^2-8a-4a-a^3+4a^2 - 8} \\ {\implies a^3-a^3+12a-8a-4a-6a^2+2a^2+4a^2+8-8} \\ {\implies a^3-a^3+12a-12a-6a^2+6a^2+8-8=0}}$

$\large {\underline {\underline {\red {Hence\:Proved\: \Rightarrow a^3+6ap+p^3-8=0}}}}$

$\rule{200} {2}$

Thank You.

Answer 2

yrr ab toh krdo plzz...