prove that if P is equal to 2 minus A then a cube + 6ap + p
cube minus 8 is equal to zero
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Answered by
3
Answer:
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.
: )
please mark it as a brainlist answer
Answered by
6
Question :
Prove that If p = 2 - a then a³ + 6ap + p³ - 8 = 0
Given,
P = 2 - a
P + a = 2
a = 2 - p ------(1)
a³ + 6ap + p³ - 8 = 0
Substitute eq - (1) in the place of a.
(2 - p)³ + 6(2 - p)p + p³ - 8 = 0
we know that,
(x - y)³ = x³ - y³ - 3x²y + 3xy²
2³ - p³ - 3(2)²(p) + 3(2)(p)² + 12p² - 6p² + p³ - 8 = 0
8 - p³ + p³ - 12p² + 6p² + 12p² - 6p² - 8 = 0
8 - 8 - p³ + p³ - 12p² + 12p² + 6p² - 6p² = 0
0 - 0 - 0 + 0 = 0
0 = 0
Hence, Proved !!
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