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Is this system of three equations solvable in arithmetic operations and root extraction etc?

$$\begin{array}{cl}x + y + z &= a\\ xyz &= b\\ x^3y^3 +x^3z^3 +z^3y^3 &=c\end{array}$$

If the degree of exponents were same , the solution will just be the factors of a cubic $$(t+x)(t+y)(t+z)=0$$

and substitution method will just result a $9$-th degree equation.

achille hui
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user781074
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1 Answers1

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Yes, $x,y,z$ are solvable using arithmetic operations and root extractions.

Let $u = \frac1x, v = \frac1y, w = \frac1z$ and define

$$\begin{align} A &= u + v + w\\ B &= uv+vw+wu = \frac{x+y+z}{xyz} = \frac{a}{b}\\ C &= uvw = \frac{1}{xyz} = \frac1b\\ p_k &= u^k + v^k + w^k\quad\text{ for } k = 1,2,3\end{align}$$ By Newton's identities, we have

$$\begin{align}p_1 &= A\\ p_2 &= Ap_1 - 2B = A^2 - 2B\\ p_3 &= Ap_2 - Bp_1 + 3C = A^3 - 3BA + 3C \end{align}$$ Notice $$p_3 = u^3 + v^3 + w^3 = \frac1{x^3} + \frac1{y^3} + \frac1{z^3} = \frac{c}{b^3}$$

$A$ is a root of the cubic equation $$t^3 - \frac{3a}{b}t + \left(\frac3b - \frac{c}{b^3}\right) = 0$$ Using Cardano's formula for depressed cubic, we can compute $A$ using arithmetic operations and root extractions from $\frac{3a}{b}$ and $\frac3b - \frac{c}{b^3}$.

Once we get $A$, $u,v,w$ are roots of the cubic equations $$t^3 - At^2 + Bt + C = 0$$ Transform the cubic equation to a depressed one and apply Cardano's formula again, we can compute $u,v,w$ (and hence $x,y,z$) using arithmetic operations and roots extractions from $A,B,C$ (and hence from $a,b,c$).

The actual formula for $u,v,w$ will be a nightmare, I will leave the fun to you.

achille hui
  • 122,701
  • I already knew the answer haha , I wanted to see a different approach . It is almost similar to my approach ,instead I find an a equation/identity for x^3+z^3 +y^3 in terms of the coefficents which results a cubic equation.Then I cube the equation for the b coefficent. Now that the exponents are same degree , the equations are just the coefficients of a cubic (t+x^3)(t+y^3)(t+z^3) :) – user781074 May 30 '20 at 20:21