If p = 2 - a ,, prove that a² + 6ap + p³ - 8 = 0
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Given:
p=2-a
To prove:
a²+6ap+p³-8=0
Proof:
RHS=0
LHS:
a²+6ap+p³-8
Put p=2-a in a²+6ap+p³-8
a²+6a(2-a)+(2-a)³-8
a²+12a-6a²+[2³-a³-3×2a(2-a)]-8
12a-6a²+8-a³-6a(2-a)-8
12a-6a²+8-a³-12a+6a²-8=0
Therefore, LHS=RHS
Hope this helps you
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