Algebraic system of equations problem

Question

Solve the follow system of equations:

$$x+y+z=5$$ $$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=5$$ $$x^3+y^3+z^3=53$$

Thanks for any help.

Answer 1

The first two equations give $xyz=1$. From the last equation we get $$x^3+y^3+z^3=(x+y+z)^3+3xyz-3(x+y+z)(xy+yz+zx)$$ and hence $xy+yz+zx=5$, so $x,y,z$ are the roots of the cubic $a^3-5a^2+5a-1=0$.

That factorises as $(a-1)(a^2-4a+1)=0$, so the roots are $1,2\pm\sqrt3$.

Answer 2

Let $z=5-x-y$. By computing the $S$-polynomials, the last two equations yield either $$ x-1=0, \; y^2 - 4y + 1=0, $$ or $$ x^2 - 4x + 1=0, \; x+y - 4=0, $$ or $$ x^2 - 4x + 1=0, \; y - 1=0. $$ Here the quadratic equation $z^2-4z+1=0$ has the two solutions $2\pm \sqrt{3}$.

Answer 3

HINT: Find $xy + yz + zx$ and $xyz$. Then use the three given equations to find $x$, $y$ and $z$ by considering the corresponding the roots of the polynomial $t^3 + \alpha t^2 + \beta t + \gamma$ where $-\alpha = x_1 + x_2 + x_3$, $\beta = x_1x_2 + x_2x_3 + x_3x_1$ and $\gamma = -x_1x_2x_3$