Algebraic Three Systems of Equations

Question

Given the three equations

\begin{align*} a+b+c&=0, \\ a^2+b^2+c^2&=2,\\ a^3+b^3+c^3&=19, \\ \end{align*}

Compute $abc.$

I know that $(a^2+b^2+c^2)^2 = (a^4+b^4+c^4)+2(a^2b^2+a^2c^2+b^2c^2),$ but this isn't leading me anywhere. Any ways to do this problem? Thanks in advance.

Answer 1

$$(a^2+b^2+c^2) = (a+b+c)^2-2(ab+bc+ca) \\ \implies ab + bc + ca = -1$$ $$(a^3+b^3+c^3) -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\ \implies 19-3abc = 0 \text{ or } abc = \frac{19}{3}$$

Answer 2

A slightly different approach: $$0=(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=2+2(ab+ac+bc)$$ $$2(ab+ac+bc)=-2$$ $$ab+ac+bc=-1$$ Similarly: $$0=(a+b+c)^3=a^3+b^3+c^3+3(\mathbf{a^2b+ab^2+a^2c+ac^2+b^2c+bc^2})+6abc$$ $$0=19+3(\mathbf{a^2b+ab^2+a^2c+ac^2+b^2c+bc^2})+6abc$$ $$-19=3(\mathbf{a^2b+ab^2+a^2c+ac^2+b^2c+bc^2})+6abc \hspace{10mm} \mathbf{(1)}$$

Now you should consider how to evaluate the part in bold by making use of what you already know ($abc$ is already in equation $\mathbf{(1)}$). You should notice by playing around with the result $ab+ac+bc=-1$ that you get the terms in bold by multiplying $$(ab+ac+bc)\cdot(a+b+c)=\mathbf{a^2b+ab^2+a^2c+ac^2+b^2c+bc^2}+3abc$$ $$-1\cdot0=\mathbf{a^2b+ab^2+a^2c+ac^2+b^2c+bc^2}+3abc$$ $$\mathbf{a^2b+ab^2+a^2c+ac^2+b^2c+bc^2}=-3abc$$

And when you plug it into $\mathbf{(1)}$ you get: $$-19=3\cdot(-3abc)+6abc$$ $$-19=-3abc$$ $$abc=\frac{19}{3}$$

Answer 3

We don't need the condition $a^2+b^2+c^2=2$.

Indeed, $$19=a^3+b^3+c^3=a^3+b^3+c^3-3abc+3abc=$$ $$=(a+b+c)\sum_{cyc}(a^2-ab)+3abc=3abc,$$ which gives $$abc=\frac{19}{3}.$$