To Know Whether Everyone Knows Identities? - 2

$a+b+c=0\\a+b=-c\\(a+b)^3=(-c)^3\\a^3+b^3+3ab(a+b)=-c^3\\a^3+b^3-3abc=-c^3\\ \text{since}\,a+b=-c\\a^3+b^3+c^3=3abc$

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

The entire right hand side is 0 because $a+b+c=0$ . So it must be that the left hand side is 0.

Hence, $a^{3}+b^{3}+c^{3}-3abc=0 \longrightarrow a^{3}+b^{3}+c^{3}=3abc$ .

Let's start with grouping the terms.

$b^{3}+c^{3}=(b+c)((b+c)^{2}-3bc)$

then we have $b+c=-a$ substitute this result to the above expression.

$b^{3}+c^{3}=(-a)(a^{2}-3bc)=-a^{3}+3abc$

So

$a^{3}+b^{3}+c^{2}=3abc$

$Let's\quad define\quad { S }_{ k }={ a }^{ k }+{ b }^{ k }+{ c }^{ k }\\ By\quad recurrence\quad we\quad got:\\ { S }_{ k+3 }-{ \sigma }_{ 1 }{ S }_{ k+2 }+{ \sigma }_{ 2 }{ S }_{ k+1 }-{ \sigma }_{ 3 }{ S }_{ k }=0\quad ,\quad k\ge 0\\ Such\quad that\quad \begin{cases} { \sigma }_{ 1 }=a+b+c \\ { \sigma }_{ 2 }=ab+ac+bc \\ { \sigma }_{ 3 }=abc \end{cases}\\ if\quad k=0\quad then\quad we\quad have:\\ \qquad \qquad { S }_{ 3 }-{ \sigma }_{ 1 }{ S }_{ 2 }+{ \sigma }_{ 2 }{ S }_{ 1 }-{ \sigma }_{ 3 }{ S }_{ 0 }=0\\ \qquad \qquad \Rightarrow { S }_{ 3 }-3{ \sigma }_{ 3 }=0\Rightarrow { S }_{ 3 }=3{ \sigma }_{ 3 }\\ Hence\quad { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=3abc\quad \Box$ note that ${ \sigma }_{ 1 }={ S }_{ 1 }=0$