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

Answers (1)

solution: we have ,

p=2-aRightarrow a+p-2=0

now   a^3+6ap+p^3-8

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

=left  a+p+(-2)^3 
ight \left  a^2+p^2+(-2)^2-ap-p(-2)-a(-2) 
ight

=(a+p-2)(a^2+p^2+4-ap+2p+2a)=0	imes(a^2+p^2+4-ap+2p+2a)=0                   [ecause a+p-2=0]

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