Question

a+2b+c=0 show that a^3+8b^3+c^3=6abc

Answer 1

We know that,

If x+y+z = 0

Then x^3 +y^3+z^3 = 3xyz

By using this identity,

(a)^3 + (2b)^3 + (c)^3 = 3 (a) (2b)(c)

=> a^3 + 8b^3 + c^3 = 6abc

Hence Proved......

Answer 2

Hope u like my process..
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Here we go,

Formula to be used:
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( x + y) ³ = x³ + y³ +3xy(x +y)
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So,, a + 2b +c =0

or, a + 2b = - c

Cubing both sides we get,
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(a + 2b) ³ = (-c) ³

or, (a)³ + (2b)³ +3.a.2b.(a +2b) = - c³

or, a³+8b³+6ab×(-c) 9+c³=0__(since a+2b =-c)

or, a³ + 8b³ +c³ - 6abc =0

or, a³ + 8b³ + c³ =6abc __[PROVED]
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Hope u got it

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