# Solving a System of 3 Equations

Let $$(a,b,c)$$ be the real solution to the system of equations: $$x^3-xyz=2$$$$y^3-xyz=6$$$$z^3-xyz=20$$ Find the greatest possibe value of $$a^3+b^3+c^3$$

My approach to the question is: I added all 3 equations and got $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=28$$ From here it can be said that $$(x+y+z)(\frac{2}{x}+\frac{6}{y}+\frac{20}{z})=28$$.

How to proceed and get the answer?

We have \begin{align} &x^3=xyz+2\\ &y^3=xyz+6\\ &z^3=xyz+20 \end{align} Denote $$t=xyz$$ and take the product of the 3 equations above, we obtain a quadratic equation of $$t$$: $$t^3 =(t+2)(t+6)(t+20)\tag{1}$$ The quadratic equation $$(1)$$ has 2 roots: $$t=-4$$ and $$t=-15/7$$.

It’s easy to check that if $$xyz=t=-15/7$$, we have a real solution $$(x,y,z)=(-1/\sqrt[3]{7},3 /\sqrt[3]{7},5 /\sqrt[3]{7} )$$ and if $$xyz=t=-4$$, we have a real solution $$(x,y,z)=(-\sqrt[3]{2},\sqrt[3]{2},2 \sqrt[3]{2} )$$

Because $$-15/7>-4$$, the maximum value of $$a^3+b^3+c^3$$ is $$2 +t + 6+t+ 20+t = 28 +3t =\color{red}{ \frac{151}{7}}$$

• Yeah Thanks for the help :) Commented Aug 1 at 17:00
• Wait why should there be no real solution for $t=-15/7$? @NN2
– Immi
Commented Aug 1 at 17:19
• @NN2 the real solution corresponding to $xyz=-15/7$ is $(x,y,z): \left(-\frac1{\sqrt[3] 7}, \frac3{\sqrt[3]7}, \frac5{\sqrt[3] 7}\right)$. Commented Aug 1 at 17:46
• @NN2: Maybe I am missing something, but doesn't the other root give a larger sum of $a^3 + b^3+c^3$? The first gives $16$ and the second $21.5714$.
– Moo
Commented Aug 1 at 18:12
• @Sahaj Thanks for pointing the mistake, I corrected it.
– NN2
Commented Aug 1 at 19:55