**If a + 2b + 3c = 0, Prove that a**^{3} + 8b^{3} + 27c^{3} = 18 abc

^{3}+ 8b

^{3}+ 27c

^{3}= 18 abc

**Answer :**

Given:

a + 2b + 3c = 0, a + 2b = – 3c

Let us cube on both the sides, we get

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

a^{3} + (2b)^{3} + 3(a) (2b) (a + 2b) = -27c^{3}

a^{3} + 8b^{3} + 6ab (– 3c) = – 27c^{3}

a^{3} + 8b^{3} – 18abc = -27c^{3}

a^{3} + 8b^{3} + 27c^{3} = 18abc

Hence proved.

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