Given the values of $abc$, $ab + ac + bc$ and $a + b + c$, how do we find the value of $a^{3} + b^{3} + c^{3}$?

Question

Given the system of equations

\begin{align*} \begin{cases} \hfil abc &=\;\; -6\\[6pt] \hfil a + b + c &=\;\; \phantom{-}2\\[6pt] \hfil ab + bc + ca &=\;\; -5 \end{cases} \end{align*}

How do I find the value of $a^3 + b^3+ c^3$?

Answer 1

According to the Vieta's formulas, the roots of the following polynomial \begin{align*} p(x) = x^{3} - 2x^{2} - 5x + 6 \end{align*} are given by $a$, $b$ and $c$. Then we have that: \begin{align*} a^{3} + b^{3} + c^{3} & = (2a^{2} + 5a - 6) + (2b^{2} + 5b - 6) + (2c^{2} + 5c - 6)\\\\ & = 2(a^{2} + b^{2} + c^{2}) + 5(a + b + c) - 18\\\\ & = 2[(a + b + c)^{2} - 2(ab + ac + bc)] + 5(a + b + c) - 18\\\\ & = 2(4 + 2\times 5) + 5\times 2 - 18 = 20 \end{align*}

Answer 2

A more cumbersome method:

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) $$$$a^3+b^3+c^3=2 \cdot(a^2+b^2+c^2-ab-bc-ca)+3\times (-6)$$$$= 2 \cdot((a+b+c)^2-3(ab+bc+ca))-18$$$$=2\cdot(4-3(-5))-18=20.$$