Question

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


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Answer 1

Answer:

a³+8b³+27c³ = 18abc

Step-by-step explanation:

a³+8b³+27c³ = a³+(+2b)³+(3c)³

Since, a + 2b + 3c =0

We have,

a³+8b³+27c³ = a³+(+2b)³+(3c)³

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

+18abc

Answer 2

a+2b+3c=0

$a^{3} + (2b)^{3} + (3c)^{3} - 18abc =$ $(a+2b+3c)(a^{2} + (2b)^{2} + (3c)^{3} + (2(a)(2b)) + (2(2b)(3c)) + (2(a)(3c)) )$

$a^{3} + (2b)^{3} + (3c)^{3} - 18abc =$ 0

$a^{3} + (2b)^{3} + (3c)^{3} = 18abc$

Solved