2
$\begingroup$

I want to solve the following three equations for $x,y,z$:

$\begin{eqnarray}k_1&=&\frac{x+y+z}{yz}\\ k_2&=&\frac{x^2+y^2+z^2}{y^2z^2}\\ k_3&=&\frac{x^3+y^3+z^3}{y^3z^3}\end{eqnarray}$

where $k_1,k_2,k_3$ are constants.

Is there any kind of standard methods?

  • 2
    You should consider accepting answers to some of your previous questions; it marks them as completed and helps reward answerers. Also, can you give us an idea of what level of math you are familiar with? For example, if I were to talk about the algebraic variety corresponding to the ideal $(x+y+z-k_1yz)$, would that make sense to you?2012-02-24

1 Answers 1

3

If you set $\frac{x}{yz} = p$, $\frac{1}{y} = q$, $\frac{1}{z} = r$, then your equations are

$p + q + r = k_1$

$p^2 + q^2 + r^2 = k_2$

$p^3 + q^3 + r^3 = k_3$

Using $(p+q+r)^2 - (p^2 + q^2 + r^2) = 2(pq + qr + rp)$ we get

$pq + qr + rp = (k_1^2 - k_2)/2$

Similarly we can find the value of $pqr$.

Thus we can find the polynomial $(x-p)(x-q)(x-r)$, by finding its coefficients. Once we have the polynomial, since it is a cubic, it is solvable by standard methods.

Once we find $p,q,r$, finding $x,y,z$ should not be difficult (keep in mind the permutations possible).

See: Newton's identities for a way to find the coefficients.

See: Roots of cubic for a formula for the roots of cubic.

  • 0
    @Prasenjit: If this did not help you, you can ask questions to clarify.2012-03-13