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The equation $$\sqrt[3]{x^2-1} + x = \sqrt{x^3-2}$$ has a solution $x = 3.$ How to solve this eqution?

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2 Answers 2

You have to rewrite this as $$ \sqrt[3]{x^2-1}= \sqrt{x^3-2}-x $$ and take the 3rd power of both members obtaining $$ x^2-1=(x^3-2)\sqrt{x^3-2}-3x(x^3-2)+3x^2\sqrt{x^3-2}-x^3. $$ Now, you can isolate the square root obtaining $$ 3x^4+x^3+x^2-6x-1=(x^3+3x^2-2)\sqrt{x^3-2}. $$ I think from here you can go on by yourself.

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Now we'll have to square it and solve a 8 degree equation? are u kiding? –  Mr.ØØ7 Apr 22 '13 at 13:04
    
After factoring out $(3-x)$ one gets $x^8+3x^6-4x^5-2x^4-7x^3+5x^2+5x+3$ which is irreducible according to maple. Maybe someone could show irreducible by Eisenstein after a change of variables... –  coffeemath Apr 22 '13 at 13:12
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@exploringnet: This is a standard route and a solution is already known. Of course, I will upvote whoever will provide an approach with a straightforward solution. –  Jon Apr 22 '13 at 14:32

The domain of the equation $$\sqrt[3]{x^2-1} + x = \sqrt{x^3-2}$$ for real $x$ comes from the squareroot, so that it is $[2^{1/3},\infty)$. Thus we may safely divide through by $x$ and write it in the form $f(x)=g(x)$ where $$f(x)=(x^{-1}+x^{-3})^{1/3}+1,\\ g(x)=(x-2x^{-2})^{1/2}.$$ One effect of the division by $x$ is that it makes it easier to analyze where the sides $f,g$ are increasing and decreasing.

A simple calculation then shows $f$ is increasing on $[2^{1/3},\sqrt{3}]$ and thereafter decreasing on $[\sqrt{3},\infty)$, and that $g$ is increasing on the entire domain $[2^{1/3},\infty).$

Between $2^{1/3}$ and $\sqrt{3}$, $g$ increases from $0$ to about $1.03217$, while in the same interval $f$ increases from about $1.6647$ to $1.7247$. This means $f$ lies completely above $g$ on the interval $[2^{1/3},\sqrt{3}]$ so that $f=g$ has no roots there.

After that, on $[\sqrt{3}, \infty)$, $f$ decreases while $g$ increases, so that there can only be one root in all of $f=g$, which as noted in the post is at $x=3$ where $f=g=5/3.$

[note that the approximated numbers have in fact rather simple exact expressions; just didn't want to clutter things up with that detail.] I'm not really that satisfied with a derivative approach, but at least after divisionj by $x$ it became tractable.

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