Systems of cubic equations

Question

I am looking for a general yet tailor-made methods to solve system of equations like that

$a^3 + 6ab^2 = 7$ and $3a^2b +2b^3 = 5$

that involve terms of $(a +b)^3$.

Does this have a name?

Answer 1

Let $\,u = a + b\sqrt{2}\,$ then $\,u^3 = a^3 + 6ab^2 + (3a^2b+2b^3) \sqrt{2}=7 + 5\sqrt{2}\,$, so the problem is equivalent to denesting $\,\sqrt[3]{5\sqrt{2} \pm 7} = b\sqrt{2} \pm a\,$.

The following steps are similar to my more general answer here.

• Let $\,v=\sqrt[3]{5 \sqrt{2} - 7}\,$, then $\,u^3-v^3=14\,$ and $\,uv = \sqrt[3]{5^2 \cdot 2 - 7^2}=1\,$.

• It follows that $\,(u-v)^3 = u^3-v^3 -3uv(u-v)=14 -3(u-v)\,$, so $\,t=u-v\,$ satisfies the cubic $\,t^3 + 3 t - 14 = 0\,$. The only real root is $\,t=2\,$, so $\,u-v=2\,$.

• Substituting $\,v=\dfrac{1}{u}\,$ in $\,u-v=2\,$ gives $\,u^2 - 2u - 1=0\,$ with roots $\,1 \pm \sqrt{2}\,$. Given that $\,u \gt 0\,$, it must be the positive root, so $\,u=\sqrt{2}+1\,$ and $\,v = \dfrac{1}{u} = \sqrt{2} - 1\,$.

It follows that $\,u,v=b\sqrt{2}\pm a=\sqrt[3]{5 \sqrt{2}\pm 7}=\sqrt{2}\pm 1\,$, so $\,a=b=1\,$.

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[ EDIT ] $\;$ If the cubic at step $2$ did not have a "nice" root like $t=2$, the problem could still be solved the same way, but the following calculations would involve the cubic formula, thus be a lot uglier, and $\,a,b\,$ would not necessarily end up being "nice" rational values.

Answer 2

Here is one way to attack a system with only degree-three terms on the left and constants on the right:

Start with your given equations:

$a^3+6ab^2=7$

$3ab^2+2b^3=5$

Take the linear combination that eliminates the constant term, in this case $5$ times the first equation minus $7$ times the second. This gives

$5a^3-21a^2b+30ab^2-14b^3=0.$

In this homogeneous equation divide by $b^3$ and define $x=a/b$ getting an equation with just $x$:

$5x^3-21x^2+30x-14=0.$

If this has a rational root, you can then plug $a=(\text{that root})b$ into either equation and solve for $b$, then get $a$ after that.

Here $x=1$, so the first equation becomes $6b^3+b^3=7$, whence $b=1$ therefore $a=1×1=1$.