Question

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

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Answer 1

Refer the attachment

Formula used :

(a + b)³ = a³ + b³ + 3ab(a + b)

#ಕನ್ನಡತಿ

Answer 2

Answer :

If a+2b+3c=0 then a³+8b³+27c³=18abc

Given :-

a +2b+3c = 0

To find :-

Prove that a³+8b³+27c³ = 18abc

Solution :-

Given that

a+2b+3c = 0

=> a +2b = -3c -------(1)

On cubing both sides then

=>(a+2b)³ = (-3c)³

=> a³+(2b)³+3(a)(2b)(a+2b) = -3³c³

Since (a+b)³ = a³+3ab(a+b)+b³

Where a = a and b = 2b

=> a³+8b³+6ab(a+2b)=-27c³

=> a³+8b³+6ab(-3c)=-27c³

=>a³+8b³-18abc = -27c³

=> a³+8b³-18abc+27c³ = 0

=> a³+8b³+27c³ = 18abc

Hence, Proved.

Answer:-

If a+2b+3c=0 then a³+8b³+27c³=18abc

Used formulae:-

• (a+b)³ = a³+b³+3ab(a+b)

More formulae to know :

• (a+b)³ = a³+b³+3a²b+3ab²

• If a +b+c = 0 then a³+b³+c³ = 3abc