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Well, I was given a problem, find x, if:


I tried everything that I could, I even opened up the brackets which gave an ugly degree 9 equation, I also tried to plot the curves $y=\left(x^2+100\right)^2$ and $y=\left(x^3-100\right)^3$ and locate their point of intersection but it couldn't be done manually.

So, in the end I was forced to use hit and trial after doing which I got the answer, is their any way to solve this algebraically??

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up vote 7 down vote accepted

For a particular solution, you can take $$x=5.$$ How did I find it? I restrict to the case of integers. We are looking for numbers such that $n^2=m^3$. Then, we expect to have $n=k^3$ and $m=k^2$ for some integer $k$. Now, we look for $$k^3=x^2+100,\quad k^2=x^3-100.$$ Those two equations would be satisfied if $x=k$, and $$k^3=k^2+100.$$ Now this equation is much simpler. I thought of what integer $k$ is such that $k^3$ is a bit above $100$ and $5^3=125$ was an obvious candidate.

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Sketching a graph of each of the curves gives you that they'll only have one intersection. – πr8 Jan 10 at 17:22
@πr8 well I tried it but i couldn't do it manually – Nikunj Jan 10 at 17:23
Couldn't draw it manually? The LHS is going to basically look like a parabola (from a distance) and the RHS like a $y=x^3$ type of curve, and looking at the powers of $x$ in each ought to hint that the RHS will be increasing faster eventually. Verify that this is the case for $x\geq5$ and you have the uniqueness. – πr8 Jan 10 at 17:29
In an examination I have 1 hr for my mathematics section, and I have 30 problems, each unique and difficult(for me), so when I tried to graph it, I got that the curves intersect at only 1 point, but I couldn't get that point. (Considering I gave this problem 3-4 minutes). – Nikunj Jan 10 at 17:34

There is a quite dirty trick you can use here. Define $$ f(x,t)=(x^2+t)^2-(x^3-t)^3\ . $$ Then solve $f(x,t)=0$ for $t$ (not for $x$). There is a simple solution of the cubic equation in $t$ $$ t=x^2(x-1)\ . $$ Then set $t=100$ and solve the cubic equation for $x$, yielding $x=5$ as the only real root.

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Nice one, although the trick that you used is pretty difficult to come by in the middle of an exam! – Nikunj Jan 10 at 17:29
But $t=x^2(x-1)$ is not the only solution. The other solutions are $t=\frac{2x^3+x^2\pm\sqrt{-3x^4 -8x^3-6x^2+1}-1}{2}$, which are real if and only if $x\in\left[-1,\frac{1}{3}\right]$ (see WolframAlpha). – user236182 Jan 11 at 0:58

It's obvious that $x^3>100$ so $x>0$ .

Consider the $6$-th root of the equation to get :


Now consider the function :$$f(x)=\sqrt[3]{x^2+100}$$

This function is bijective from $(0,\infty$) to $(\sqrt[3]{100},\infty)$ .

Its inverse is :$$f^{-1}(x)=\sqrt{x^3-100}$$

This means that the equation is now :



But $f$ is an increasing function so let's take two cases :

  • If $f(x)>x$ then :

$$x=f(f(x))>f(x)>x$$ a contradiction .

  • If $f(x)<x$ then : $$x=f(f(x))<f(x)<x$$ a contradiction .

This means that $f(x)=x$ so : $$x^3=100+x^2$$ which can be solved easily .

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The best solution so far. – whatever Jan 10 at 19:03
@whatever Thanks . I actually know this trick from some time so when I saw this problem it immediately popped to me that it might work , and it did . – ComplexPhi Jan 10 at 19:09

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