Question

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

Answer 1

hope it helps.........

Answer 2

Answer:

Taking RHS

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

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

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

a^3 - a^3 + 12a - 12a - 6a^2 + 6a^2 + 8 - 8 =

0

OR

You can do the question by LHS also,vo jada simple rhega

given p=a-2 then a+p=2

(a+p)^3 = 8

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

a^3+3ap.(2)+p^3-8 = 0

a^3+6ap+p^3-8 = 0

Hope so you got your answer :))