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

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

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Given `p=2−a`
to prove `a^3+6ap+p^3−8=0`...(1)
Now put the value of `P` in eq (1)
`a^3+6a(2−a)+(2−a)^3−8=0`
`=>a^3+12a−6a^2+8−a^3−12a+6a^2−8=0`
`[∴(a−b)^3=a^3−b^3−3a^2b+3ab^2]`
`=>0=0` (By eliminating equal terms we have LHS = RHS)
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