Question

If p=2-a, Solve:a^3+6ap+p^3-8

Answer 1

Hi ,

It is given that ,

p = 2 - a -------( 1 )

LHS = a³ + 6ap + p³ - 8

= a³ + 6a( 2 - a ) + ( 2-a )³ - 8 [ from(1)]

= a³ +12a-6a²+2³-3×2²a+3×2a²-a³-8

= a³ + 12a - 6a² + 8 - 12a + 6a²- a³ - 8

= 0

= RHS

Hence proved.

I hope this helps you.

: )

Answer 2

a³+6ap+p³-8
putting p=(2-a)

a³+6a(2-a)+(2-a) ³-8
=a³+12a-6a²+8-a³-12a+6a²-8
=0