Question

if a+b+c=0then show that acube+bcube+cube=3abc

Answer 1

✍✍HERE IS YOUR SOLUTION…⬇⬇
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$\underline{SOLUTION}$ ➡

$\underline{Given : - }\: \: a + b + c = 0$

$\underline{To \: prove : -} \: \: a {}^{3} + b {}^{3} + c {}^{3} = 3abc$

$\underline{Proof : -}$

$We \: \: know \: \: that, \\ \\ a {}^{3} + b {}^{3} + c {}^{3} - 3 abc =( a+ b + c)\\ (a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) - - - > (1)\\ \\ \\ Putting \: \: the \: \: value \: of \: \: a + b + c = 0 \: \:\\ in \: \: (1), \: \: \: We \: \: get ,\\ \\ \\ a {}^{3} + b {}^{3} + c {}^{3} - 3abc = 0 \times\\ (a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) \\ \\ = > a {}^{3} + b {}^{3} + c {}^{3} - 3abc = 0 \\ \\ = > a {}^{3} + b {}^{3} + c {}^{3} = 3abc$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Hence \: \: Proved \: .$

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Answer 2

Answer:

Step-by-step explanation:

Given:a+b+c=0

To prove:a3+b3+c3=3abc

Proof:a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca) (here 2 means square)

Taking RHS

(a+b+c)(a2+b2+c2-ab-bc-ca)

(0)(a2+b2+c2-ab-bc-ca)

0

Therefore,a3+b3+c3-3abc=0

a3+b3+c3=3abc