Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known.

Question

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?

Answer 1

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

Answer 2

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 $$\color{red}{a^3+b^3+c^3=(a+b+c)^3+3abc-3(a+b+c)(ac+ab+bc)}$$

Answer 3

We have $$ (a + b + c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab + bc + ca) - 3abc $$

Answer 4

Yes, there is. Hints:

1. 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.
2. Use the expansion of $(a+b+c)(ab+bc+ca)$ to cancel most of those terms.
3. Some additional term still remaining? Use $abc$.

Answer 5

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