Math, asked by vaishnavi8643, 1 year ago

prove that if P is equal to 2 minus A then a cube + 6ap + p
cube minus 8 is equal to zero​

Answers

Answered by harsh427868
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 CaptainBrainly
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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