Solve the system $a^3+b^3= 28$, $(a+2)(b+2)(a+b)= 60$

Question

Solve the system $a^3+b^3= 28$, $(a+2)(b+2)(a+b)= 60$, where $a,b \in \mathbb{R} $.

My try:

If $a,b \in \mathbb{N}$:

$$(a+2)(b+2)(a+b)=3\times 4\times 5 \to a=1, \, b=3 $$

$$a^3+b^3=27+1 \to a=1, \, b=3$$

but $a,b \in \mathbb{R}$

Answer 1

Multiply the second equation by 3 and add it to the first. Then $$a^3+b^3+3(a+2)(b+2)(a+b)=208\\ a^3+b^3+3(ab+2a+2b+4)(a+b)=208\\ (a+b)^3+6(a+b)^2+12(a+b)=208$$ Let $a+b=x$. Then $x^3+6x^2+12x=208$. Since you already know one root, you're left with a quadratic equation.

Answer 2

Rewriting the equations in terms of $s=a+b, \,p=ab\,$ gives:

$$ \begin{cases} \begin{align*} s^3 - 3 p s &= 28 \\ (p+2s+4)s &= 60 \end{align*} \end{cases} $$

Eliminating $p$ between the two equations results in a cubic in $s$: $s^3+6s^2+12s-208=0\,$. Trying out factors of $208=2^4 \cdot 13$ finds the root $s=4$, which then gives $p=3$ so $\,\{a,b\}=\{1,3\}\,$.

Factoring out $s-4$ leaves a quadratic $\,s^2 + 10 s + 52 = 0\,$ which gives a couple of complex roots $s=-5 \pm 3\sqrt{3} i$, each one corresponding to a pair of complex $a,b$.

Answer 3

Since the equations are symmetric in $a$ and $b$ we obtain with $(a,b)=(1,3)$ also $(a,b)=(3,1)$ as a solution. Indeed, computing the resultant of the two polynomials we obtain $$ (b-1)(b-3)(13b^4+130b^3+682b^2+2244b+3492)=0. $$ The last term has no real solution. So we obtain the above real solutions. However, we obtain $4$ new complex solutions.