I want to sove the system of equations $$\begin{cases} x^3 y-y^4=7,\\ x^2 y+2 xy^2+y^3=9. \end{cases} $$ I tried divide these two equations we obtain $$\dfrac{x^3 - y^3}{(x+y)^2 } = \dfrac{7}{9}$$ From here, I don't know how to solve.

  • $\begingroup$ Here's one try: $x^3-y^3=(x-y)(x^2+xy+y^2)$. $\endgroup$ Dec 20, 2012 at 4:40

3 Answers 3


We'll prove that $(2,1)$ is the only real solution. First note that if $y = 0$, the second equation is unsolvable, so $y\neq 0$.

Solving the second equation for $x$ via the quadratic formula gives \begin{align*} x &= \frac{-2y^2 \pm \sqrt{4y^4-4y(y^3-9)}}{2y}\\ &= \frac{-y^2\pm 3\sqrt{y}}{y} \\ &= -y \pm y^{-\frac{1}{2}}.\end{align*}

In particular, $y \geq 0$. Subbing this into the first equation gives \begin{align*}7 &= \left(-y \pm 3y^{-\frac{1}{2}} \right)^3 y - y^4 \\ &= -2y^4\pm 9y^{\frac{5}{2}}-27y\pm 27y^{-\frac{1}{2}}.\end{align*}

This is equivalent to $$ 0 = -2y^{\frac{9}{2}} \pm 9y^{\frac{6}{2}}-27y^{\frac{3}{2}} -7y^\frac{1}{2} \pm 27.$$

Now, substitute in $z = \sqrt{y}$, which is possible since we already know $y\geq 0$. This gives $$0 = -2z^9 \pm 9z^6 - 27z^3 -7z \pm 27$$

If we choose the $-$ sign, then clearly there is no solution with $z$ real, so we must have $$0 = -2z^9 + 9z^6 - 27z^3-7z+27.$$

Now, using the rational roots theorem, maple, direct inspection, etc, we see $z=1$ is a solution.

I claim that $f(z)-2z^9 + 9z^6 - 27z^3 - 7z + 27$ is always decreasing, so $z=1$ is the only solution.

To see this, compute the derivative, getting $f'(z)=-18z^8 + 54z^5-81z^2 - 7 = 9(-2z^8 + 6z^5-9z^2 - \frac{7}{9})$. I claim this is always negative. The $9$ out in front doesn't effect this, so we'll ignore it in intermediate computations.

Why is $f'(z)$ always negative? Lets find its maximum. So, take another derivative, getting $f''(z)-16z^7+30z^4-18z = 2z(-8z^6 + 15z^3 - 9)$. Of course, this is $0$ when $z=0$, but the other part is quadratic in $z^3$ and so we see that it's never $0$. Since $f'(z)$ goes to $-\infty$ as $x\rightarrow \pm \infty$, it follows that if it only has one critical point, this must be a max. So, the max of $f'(z)$ occurs when $z = 0$, where $f'(0) = -7 < 0$.

Since $f'(z) < 0$, $f(z)$ is always decreasing, so $z= 1$ must be its only solution.

When $z=1$, $\sqrt{y} = 1$ so $y=1$, and subbing back in shows that $x=2$.

  • $\begingroup$ Please check your solution at "Solving the second equation for x via the quadratic formula gives..." $\endgroup$ Dec 20, 2012 at 9:34
  • $\begingroup$ @I'm not sure what you're referring to, it still looks ok to me. $\endgroup$ Dec 20, 2012 at 14:47
  • $\begingroup$ Note that, $$\sqrt{4y^4 -4y(y^3 -9)}=6\sqrt{y}$$. $\endgroup$ Dec 20, 2012 at 15:07
  • $\begingroup$ You're right!! Let me see how this messes with everything... $\endgroup$ Dec 20, 2012 at 15:17
  • $\begingroup$ It took forever, but it's fixed (I think). $\endgroup$ Dec 20, 2012 at 16:30

Guess and check works: $x=2$, $y=1$. (I tried forcing $x+y$ to be 3 and that's what I got.)

For a more general method for finding all solutions, I suppose applying the appropriate root-finding algorithm to approximate the solutions could work.

  • 3
    $\begingroup$ Well, that's one solution. Maybe there are others? $\endgroup$ Dec 20, 2012 at 4:37
  • 1
    $\begingroup$ Alpha finds only that real root. $\endgroup$ Dec 20, 2012 at 4:48
  • $\begingroup$ Huh. Guess I got pretty lucky then. $\endgroup$ Dec 20, 2012 at 4:50
  • $\begingroup$ Depends on where the problem comes from. Class problems often have solutions you can find that way. The rational root theorem is a great example-it works better in class than in real life. This looks like one that was supposed to be solved the way you did. Looking for small integer solutions is often productive. $\endgroup$ Dec 20, 2012 at 5:47

$$\begin{cases} x^3 y-y^4=7,\\ x^2 y+2 xy^2+y^3=9. \end{cases}$$

$x^3y-y^4=y(x^3-y^3)$, while $x^2y+2xy^2+y^3=y(x+y)^3$.

We try the transformation $x+y=z$.

Then $y(x^3-y^3)=y(x-y)(x^2+xy+y^2)=y(z-2y)(z^2-xy)=y(z-2y)(z^2-(z-y)y)=$ $-2y^4+3y^3z-3y^2z^2+yz^3=7$ On the other hand, the second equation gives $yz^2=9$, so $y=\frac{9}{z^2}$. We expand and then notice that it is actually equal to $9(z^3-18)(z^6-9z^3+81)=7z^8$. Then we get $z=3$ as a root.

The problem is that I do not know how do I show that there are no other roots, but probably someone who knows calculus can be able to resolve this? or probably by a weird factoring?


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