Find the value of $a^4+b^4+c^4$

Question

The problem:

The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?

Based on the above information, we have: \begin{align} a + b + c &= 6 \\ a^2 + b^2 + c^2 & = 8 \\ a^3 + b^3 + c^3 & = 5 \\ \end{align}

I had a feeling that this vaguely had to do with Viete's theorem, which states for a cubic polynomial $f(x) = x^3 - px^2 + qx - r$ which has roots $\alpha , \beta , \gamma$,

\begin{align} p & = \alpha+\beta+\gamma \\ q & = \alpha \beta+\alpha\gamma+\beta\gamma \\ r & = \alpha\beta\gamma \end{align}

Notice that we already have $p$, because $a+b+c=6=\alpha + \beta + \gamma = p$. Then to find $q$:

\begin{align} 2q& = 2\alpha\beta+2\alpha\gamma+2\beta\gamma \\ & = [(a+b+c)^2-(a^2+b^2+c^2)] \\ & = (a^2+ab+ac+b^2+ab+bc+c^2+ac+bc)-a^2-b^2-c^2 \\ & = 2ab+2ac+2bc \end{align}

\begin{align} q & = ab+ab+bc \\ & = \frac{1}{2}[(a+b+c)^2-(a^2+b^2+c^2)] \\ & = \frac{1}{2}[6^2-8] \\ & = 14 \end{align}

So now we have $f(x) = x^3-6x^2+14x-r$. And it follows that $f(\alpha)=f(\beta)=f(\gamma)=0$.

\begin{align} f(\alpha) & = \alpha^3-6\alpha^2+14\alpha-r = 0\\ f(\beta) & = \beta^3-6\beta^2+14\beta-r = 0\\ f(\gamma) & = \gamma^3-6\gamma^2+14\gamma-r = 0\\ \end{align}

\begin{align} 0 & = f(\alpha) + f(\beta) + f(\gamma) \\ & = (\alpha^3+\beta^3+\gamma^3)-6(\alpha^2+\beta^2+\gamma^2)+14(\alpha+\beta+\gamma)-3r\\ & = 5-6(8)+14(6)-3r\\ 3r& = 41\\ r & = \frac{41}{3} \\ \end{align}

Now we have that $f(x)=x^3-6x^2+14x-\frac{41}{3}$. Here's where I am stuck. Of course, the above working out was the culmination of hours of trying things out, and eventually we have this equation. If you are reading this now, I'd appreciate if you gave any hints as to how I should continue the problem.

I've had the idea of multiplying $f(x)$ by $x$ to get fourth powers, but I haven't tried that yet. Perhaps that might yield some results?

Answer 1

$f(\alpha)=0$ implies that $\alpha f(\alpha)=0$.

So, $\alpha^4-6\alpha^3+14\alpha^2-\frac{41}{3}\alpha=0$

We have similar results for $\beta$ and $\gamma$.

So, $\alpha^4+\beta^4+\gamma^4=6(\alpha^3+\beta^3+\gamma^3)-14(\alpha^2+\beta^2+\gamma^2)+\frac{41}{3}(\alpha+\beta+\gamma)$

Answer 2

Hint:

$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\iff ab+bc+ca=?$$

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

Now $$a^4+b^4+c^4=(\underbrace{a^2+b^2+c^2})^2-2(a^2b^2+b^2c^2+c^2a^2)$$

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

Answer 3

Let set $\begin{cases}S_1=a+b+c=6\\S_2=a^2+b^2+c^2=8,\\S_3=a^3+b^3+c^3=5\end{cases}$

$S_4=\frac 16\left({S_1}^4+8\,S_3\,S_1-6\,S_2\,{S_1}^2+3\,{S_2}^2\right)=0$

I found the relation by removing terms successively, if you are interested, here were my steps...

$A=S_4-S_3S_1\\ B=S_2S_1^2-S_4\\ C=2A+B\\ D=S_1^4-6C\\ E=D+3S_2^2-4S_4\\ F=E-4S_3S_1+4S_4=0$

---

If I develop lab bhattacharjee's answer with my notations we have:

$ab+bc+ca = \frac 12({S_1}^2-S_2)$

$abc = \frac 16({S_1}^3-3S_1S_2+2S_3)$

$a^2b^2+b^2c^2+c^2a^2=\frac 1{12}(-{S_1}^4+6{S_1}^2S_2+3{S_2}^2-8S_1S_3)$

And we arrive finally at the same relation $6{S_2}^2-6S_4=-{S_1}^4+6{S_1}^2S_2+3{S_2}^2-8S_1S_3$