Question

If a+b+c=0, then prove that
${a}^{3} + {b}^{3} + {c}^{3} = 3abc$
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Answer 1

Answer:

Step-bay-step explanation:

We know that

a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

= 0 *(a^2+b^2+c^2-ab-bc-ca)

= 0

=> a^3+b^3+c^3 = 3 abc. Proved