Question

If p = 2 - a ,, prove that a² + 6ap + p³ - 8 = 0

Plz Help....​

Answer 1

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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Answer 2

Step-by-step explanation:

a²+6a(2-a)+(2-a)³-8=0

a² +12a-6à²+(2)³-(a)³-3(2)²(a)+3(2)(a)².

(using identity in (2-a)³= x³-y³=x³-y³-3x²y+3xy²)

a²+12a-6a²+8-a³-12a+6a²

-a³+a²-6a²+6a²+12a-12a+8

-a³+a²+0+0+8

changing the sign off -a³ to +a³

a²=8

a=√8