If p= 2-a, prove that a3 + 6ap +p3 - 8=0
Answers
Answered by
832
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.
: )
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.
: )
Answered by
218
Answer:0=0
Step-by-step explanation:
P=2-a
=a^3+6ap+p^3-8
=a^3+6a(2-a)+(2-a)^3-8
=a^3+12a-6a^2+8-12a+6a^2-a^3-8 ...(1)
RHS=0
LHS=.....(1)
BY CANCELLATION
LHS=0
RHS=0
IMPLIES RHS=LHS
HENCE PROVED
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