Question

If a+b+c=0 then show that a^3+b^3+c^3=0

Answer 1

We know that,

$\displaystyle\longrightarrow\sf{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}$

Since $\displaystyle\sf{a+b+c=0,}$

$\displaystyle\longrightarrow\sf{a^3+b^3+c^3-3abc=0(a^2+b^2+c^2-ab-bc-ac)}$

$\displaystyle\longrightarrow\sf{a^3+b^3+c^3-3abc=0}$

$\displaystyle\longrightarrow\sf{a^3+b^3+c^3=3abc}$

$\displaystyle\sf{\therefore\ a+b+c=0}$ does not always imply $\displaystyle\sf{a^3+b^3+c^3=0}$ but implies $\displaystyle\sf{a^3+b^3+c^3=3abc.}$

As a special case, $\displaystyle\sf{a^3+b^3+c^3=0}$ if and only if at least one among $\displaystyle\sf{a,\ b}$ and $\displaystyle\sf{c}$ is zero.

QED

Answer 2

Answer:

If a+b+c=0, then a^3+b^3+c^3=3abc not 0.

Step-by-step explanation:

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

If a+b+c=0,

a^3+b^3+c^3-3abc=(0)(a^2+b^2+c^2-ab-bc-ca)

=>a^3+b^3+c^3-3abc=0

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