<s>[INST] Solving $x+x^3=5$ without using the cubic equation. In lessons, I get quite bored and recently throughout these lessons I have been trying to solve for x in:
$$x+x^3=5$$
I've figured out how to do it for squares using the quadratic equation, but the cubic equation looks so dauntingly massive it actually makes my bladder hurt.
So, is there a way to figure this out using a different process, and better so for $x^n$.
Danke Chien
 [/INST] A trick for solving this equation goes back to Viète (but is not much different from Tartaglia's and Cardano's method).
Set $x=y-\dfrac{1}{3y}$. Then
$$
x^3+x=y^3-y+\frac{1}{3y}-\frac{1}{27y^3}+y-\frac{1}{3y}=
y^3-\frac{1}{27y^3}
$$
so your equation becomes, after setting $z=y^3$,
$$
z-\frac{1}{27z}=5
$$
This becomes a quadratic:
$$
27z^2-135z-1=0
$$
and so
$$
z=\frac{135\pm\sqrt{18333}}{54}=\frac{45\pm\sqrt{2037}}{18}
$$
Let's take the root with $+$; then
$$
y=\sqrt[3]{\frac{45+\sqrt{2037}}{18}}
$$
and so
$$
x=\sqrt[3]{\frac{45+\sqrt{2037}}{18}}-
\frac{1}{3}\sqrt[3]{\frac{18}{45+\sqrt{2037}}}\approx 1.51598
$$
You can check that taking the other root yields the same solution.
In case the equation is $x^3+px=q$, the substitution is
$$
x=y-\frac{p}{3y}
$$
because in this case we get
$$
x^3+px=y^3-py+\frac{p^2}{3y}-\frac{p^3}{27y^3}+py-\frac{p^2}{3y}
=y^3-\frac{p^3}{27y^3}
$$
Setting $z=y^3$, we get the quadratic
$$
27z^2-27qz-p^3=0
$$
which has real roots if and only if
$$
27^2q^2+4\cdot27p^3\ge0
$$
that can be written
$$
\frac{q^2}{4}+\frac{p^3}{27}\ge0
$$
It turns out that the equation $x^3+px=q$ has just one real root when $q^2/4+p^3/27>0$ and a double (or triple) root when $q^2/4+p^3/27=0$.
What happens when $q^2/4+p^3/27<0$? That's the big question: it is the case when the equation has three distinct roots. The method still works, but it necessarily involves the complex numbers, if an algebraic expression of the roots is needed.
Viète devised a trigonometric method for getting approximate values of the roots in the case $q^2/4+p^3/27<0$.
OK, let's try another one: $x^3+x=2$. The quadratic becomes
$$
27z^2-54z-1=0
$$
which has a root
$$
z=\frac{9+\sqrt{84}}{9}
$$
and so
$$
x=
\sqrt[3]{\frac{9+\sqrt{84}}{9}}-
\frac{1}{3}\sqrt[3]{\frac{9}{9+\sqrt{84}}}
$$
which is just a complicated way to write $1$.
</s>