Given that $a+b+c=0$ and $a^3+b^3+c^3=3abc$, evaluate $(a-b)^3+(b-c)^3+(c-a)^3$.

Question

Accepted Answer

0. Using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, the condition $a+b+c=0$ and $a^3+b^3+c^3=3abc$ are consistent. Now let $x=a-b$, $y=b-c$, and $z=c-a$; then $x+y+z=0$, so $x^3+y^3+z^3=3xyz$. Since $xyz=(a-b)(b-c)(c-a)$ and the given expression is exactly this sum, it equals $3(a-b)(b-c)(c-a)$ only if the variables are the differences; however the standard identity for $(a-b)^3+(b-c)^3+(c-a)^3$ gives $3(a-b)(b-c)(c-a)$.

Given that $a+b+c=0$ and $a^3+b^3+c^3=3abc$, evaluate $(a-b)^3+(b-c)^3+(c-a)^3$.

Solution

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