Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known. Is there a way to to find out what $a^3+b^3+c^3$ evaluates to, when the values of $abc$, $ab+bc+ca$ and $a+b+c$ are given?
Alternatively, is there a way to express $a^3+b^3+c^3$ in terms of the aforementioned expressions?
 A: Use Newton's identities:
\begin{align}
  p_1 &= e_1\\
  p_2 &= e_1p_1-2e_2\\
  p_3 &= e_1p_2 - e_2p_1 + 3e_3 \\
\end{align}
where
\begin{align}
  e_1 &= a+b+c    &\qquad p_1 &= a^1+b^1+c^1\\
  e_2 &= ab+bc+ca &\qquad p_2 &= a^2+b^2+c^2\\
  e_3 &=abc       &\qquad p_3 &= a^3+b^3+c^3\\
\end{align}
Newton's identities give
$$
p_3 = e_1^3-3e_1e_2+3e_3
$$
that is
$$
a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc
$$
A: Yes, indeed,
$$(a+b+c)^3$$
$$=a^3+b^3+c^3+3a^2b+3a^2c+3b^2a+3b^2c+3c^2a+3c^2b+6abc$$
$$=a^3+b^3+c^3+3a^2b+3a^2c+3abc+3b^2a+3b^2c+3abc+3c^2a+3c^2b+3abc-3abc$$
$$=a^3+b^3+c^3+3a(ac+ab+bc)+3b(ac+ab+bc)+3c(ac+ab+bc)-3abc$$
$$=a^3+b^3+c^3+3(a+b+c)(ac+ab+bc)-3abc$$
So
$$\color{red}{a^3+b^3+c^3=(a+b+c)^3+3abc-3(a+b+c)(ac+ab+bc)}$$
A: We have
$$
(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab + bc + ca) - 3abc 
$$
A: Yes, there is. Hints:


*

*You need the cubes of $a$, $b$ and $c$. Try expanding $(a+b+c)^3$ and see what terms you need to cancel from there.

*Use the expansion of $(a+b+c)(ab+bc+ca)$ to cancel most of those terms.

*Some additional term still remaining? Use $abc$.

A: The answer is yes and it can be generalized as the The fundamental theorem of symmetric polynomials.
