if p=2-a,prove that a³+6ap+p³-8=0
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Answered by
8
Answer:
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
2
b+3ab
2
]
⇒0=0 (By eliminating equal terms we
have LHS = RHS)
R
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