If q = 2-a Prove that a2 + 6aq +q3-8 = 0
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Given p=2−a
to prove a3+6ap+p3−8=0...(1)
Now put the value of P in eq (1)
a3+6a(2−a)+(2−a)3−8=0
⇒a3+12a−6a2+8−a3−12a+6a2−8=0
[∴(a−b)3=a3−b3−3a2b+3ab2]
⇒0=0 (By eliminating equal terms we
have LHS = RHS)
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