Question

If a+b+c= 0 Then Prove that (a+b)*3 + (b+c)*3 +
(C+a)*3 = 3ab(a+b)​

Answer 1

We know that,

$\sf{ {a}^{3} + { b}^{3} + {c}^{3} - 3abc = (a + b +c )(a^2+b^2+c^2−ab−bc−ca)}$

Now, putting (a + b +c) = 0 on RHS

We get,

$\sf{ \implies {a}^{3} + { b}^{3} + {c}^{3} - 3abc = (0)(a^2+b^2+c^2−ab−bc−ca)}$

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

$\implies\sf \large \green { {a}^{3} + { b}^{3} + {c}^{3} = 3abc}$

Note :

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Answer 2

We know that,

\sf{ {a}^{3} + { b}^{3} + {c}^{3} - 3abc = (a + b +c )(a^2+b^2+c^2−ab−bc−ca)}a

3

+b

3

+c

3

−3abc=(a+b+c)(a

2

+b

2

+c

2

−ab−bc−ca)

Now, putting (a + b +c) = 0 on RHS

We get,

\sf{ \implies {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

2

+b

2

+c

2

−ab−bc−ca)

\sf{ \implies {a}^{3} + { b}^{3} + {c}^{3} - 3abc = 0}⟹a

3

+b

3

+c

3

−3abc=0

\implies\sf \large \green { {a}^{3} + { b}^{3} + {c}^{3} = 3abc}⟹a

3

+b

3

+c