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Solve for real $x$: $$x^3-3x^2-x=\sqrt2$$

This question was asked here $3$ hours ago and it got several solutions too, but there was a comment that pointed out a peculiar method of solving this question. I've been trying to solve the problem by that method, but unfortunately I am unable to continue with it. The comment was made by Duong Ngo and the question is this.

Let $x=u+1$, then we have $$ u^3-4u=3+\sqrt2$$ Let $u=\frac{4t}{\sqrt3}$, then we have $$4t^3-3t=\frac{3\sqrt3(3+\sqrt2)}{16}$$ Let $t=\frac12\left(a-\frac{1}{a}\right)$, then we have $$\frac{a^3}{2}-\frac{1}{2a^3}-3a+\frac{3}{a}=\frac{3\sqrt3(3+\sqrt2)}{16}$$

Here, I am incompetent to transform the above equation to a quadratic equation. Can anyone enlighten me on this?

Any help is greatly appreciated.

Blue
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Blue Cat Blues
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1 Answers1

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I guess that there is a small mistake in the comment and that the intended substitution should be $$t=\frac12\left(a+\frac{1}{a}\right)$$ from which $$4t^3-3t=\frac{1 + a^6}{2 a^3},$$ and now you get the quadratic equation in $z=a^3$.

b00n heT
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  • Thanks a lot. I have one more question tho. How did he think of these chain of substitutions? Is there a logic behind this or does it comes from sheer experience?? – Blue Cat Blues Jan 13 '25 at 15:31
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    I'm not sure, but I think this is the idea behind the substitution https://en.wikipedia.org/wiki/Cubic_equation#Vieta's_substitution as indeed it matches up to a constant. – b00n heT Jan 13 '25 at 15:35