Question

p=2-a prove that a³+6ap+p³-8=0

Answer 1

Answer:

p=2-a

p-2+a=0

p^3+(-2)^3+a^3=3(p)(-2)(a)

[[This is an identity if a+b+c=0

then a^3+b^3+c^3=3abc]]

a^3+6ap+p^3-8=a^3+p^3-2^3+6ap

=3(p)(-2)a+6ap

=-6ap+6ap=0