Question

if a+2b+3c=0 then prove that a³+8b³+27c³=18abc

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Answer 1

Answer:

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if a+2b+3c=0 then prove that a³+8b³+27c³=18abc.

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Given:-

• a + 2b + 3c = 0

=> a + 2b = -3c

Solution:-

By cubing both the sides

We get,

( a + 2b )³= ( −3c )³

• By Using Identity:- (a+b)³ = a³ + b³ + 3ab(a+b)

=> a³ + (2b)³ + 3(a)(2b)(a+2b) = -27c³

• Putting value of a + 2b = -3

=> a³ + 8b³ + 6ab ( −3) = −27c³

=> a³ + 8b³ −18ab = −27c³

=> a³ + 8b³ + 27c³ = 18ab

Hence Proved..

Answer 2

Refer the attachment.

Formula used :

(a + b)³ = a³ + b³ + 3ab(a + b)