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I am looking for ways of solving systems like that:

$$\left\{\begin{array}{lcl} a^3 - 33 ab^2 = -217 \\ 3a^2 b - 11b^3 = 18 \end{array} \right.$$

I've tried turning it into a system of 2 equations with 4 variables but then, having the relations and 4 degrees of freedom makes it equally difficult. Are there any catches one could apply here?

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3 Answers 3

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Does this help? $$(a-\sqrt{11} i b)^3 = a^3-3\sqrt{11}i a^2 b -33ab^2+11\sqrt{11}ib^3$$ $$=(a^3 -33ab^2)-\sqrt{11}i(3 a^2 b-11b^3)$$ $$=-217-\sqrt{11} \, 18 \, i$$

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  • $\begingroup$ Interesting. I would instead have focused on $\displaystyle ~(a + \sqrt{11} i b)^3.$ However, I haven't really thought the matter through, so, far all I know, your approach may be better. $\endgroup$ Commented Jun 10, 2022 at 10:39
  • $\begingroup$ It looks to me like the two approaches correspond on making a sign change on $b.$ $\endgroup$
    – coffeemath
    Commented Jun 10, 2022 at 10:40
  • $\begingroup$ Did you mean $\sqrt[3]{11}$? $\endgroup$
    – Robert B.
    Commented Jun 10, 2022 at 10:42
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    $\begingroup$ $\displaystyle r^3e^{i(3\theta)} = z^3 \implies z = re^{i\theta}.$ $\endgroup$ Commented Jun 10, 2022 at 10:43
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    $\begingroup$ @RobertB. No, that is not what he intends. See my last comment, and think about DeMoivre's Theorem, that $~\displaystyle \left[e^{i\theta}\right]^n = e^{i n\theta}.$ $\endgroup$ Commented Jun 10, 2022 at 10:45
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This is one of the other approaches you would like to try.

Multiply the first equation by $18$, multiply the second one by $217$, then add them together, we get:

$18a^3-18*33ab^2+217*3a^2b-217*11b^3=0$

$\Leftrightarrow 18a^3+651a^2b-594ab^2-2387b^3=0$

The last equation is homogeneous of degree 3, so you could solve the associate cubic equation to compute $a$ by $b$ (Luckily this cubic equation has a rational root).

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  • $\begingroup$ I struggle to see why is it 3-homogenous. Can you please elaborate? $\endgroup$
    – Robert B.
    Commented Jun 11, 2022 at 22:22
  • $\begingroup$ @RobertB. You just look at the terms of the LHS. 18a^3 has degree 3, 651a^2b has degree 2+1=3, 594ab^2 has degree 1+2=3, and 2387b^3 has degree 3 too. So the LHS is 3-homogeneous. Ah, and to find the associate cubic equation, all you have to do is just divide two side by b. Of course, remember the case b=0 before dividing :D $\endgroup$ Commented Jun 12, 2022 at 7:19
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Taking the resultant of $p=a^3-33ab^2+217$ and $q=3a^2b-11b^3-18$ with respect to $b$ immediately gives $$ (4a^4 - 14a^3 + 111a^2 + 217a + 961)(4a^2 + 14a + 49)(2a^2 + 7a - 31)(2a - 7)=0. $$ Hence the only rational solution is $$ (a,b)=\left( \frac{7}{2},\frac{3}{2}\right). $$ The quadratic equation $2a^2+7a-31=0$ yields $$ (a,b)=\left( \frac{-7\pm 3\sqrt{33}}{2},\frac{-33\pm 7\sqrt{33}}{2}\right) $$ Similarly we can solve the quadratic equation $4a^2 + 14a + 49=0$, which has non-real solutions. The quartic equation has four non-real roots.

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  • $\begingroup$ Can you please elaborate how do you get $b$? $\endgroup$
    – Robert B.
    Commented Jun 10, 2022 at 13:46
  • $\begingroup$ Actually, already $${\rm resultant}(a^3-33ab^2+217,3a^2b-11b^3-18,a)=(484b^4 - 726b^3 + 1221b^2 + 198b + 36)(22b^2 + 33b - 6)(4b^2 + 6b + 9)(2b - 3)=0,$$ so that we are done immediately. And taking the resultant of these with respect to $b$ gives exactly the condition in my answer for $a$. $\endgroup$ Commented Jun 10, 2022 at 13:53
  • $\begingroup$ Thanks, the problem is that it is hard to arrive at the factorised form of the resultant. Did you use e.g. Wolframalpha to compute it? $\endgroup$
    – Robert B.
    Commented Jun 11, 2022 at 16:29
  • $\begingroup$ Yes, I did not compute this by hand. Systems of cubic equations usually require a CAS. By the way, by the rational root theorem it is not hard to find the rational root. $\endgroup$ Commented Jun 11, 2022 at 16:50
  • $\begingroup$ Sure, but once there are non-rational roots we are doomed. So this methods can't really work without a computer. $\endgroup$
    – Robert B.
    Commented Jun 11, 2022 at 16:57

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