Prove the following

Question:

If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}$ is equal to

(a) 0

(b) $a b c$

(c) $3 a b c$

(d) $2 \mathrm{abc}$

Solution:

(d) Now, a3+b3 + c3= (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc

[using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given]

a3+b3 + c3 = 3abc

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