Question

a^3 + b^3 + c^3 – 3abc =​

Answer 1

Answer:

{a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0

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Answer 2

Answer:

Considering that:-

(a+b+c)^3−(a^3+b^3+c^3−3abc) =3(a+b+c)(ab+bc+ac)

then if (a+b+c)≠0

we have

(a+b+c)^2 = 3(ab+bc+ac)

and finally

a+b+c = +/- √3√(ab+bc+ac)

Note:-

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

=a^2+b^2+c^2 -(ab+bc+ac)

= 1/2 [ (a-b)^2 + (b-c)^2 + (a-c)^2 ]

= 0

=RHS