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
 A: See that we have the beautiful trigonometric identity
$$\sinh(3\operatorname{arcsinh}(r))=4r^3+3r$$
Which can be used here.
To give a definition of the above,
$$\sinh(r)=\frac{e^r-e^{-r}}2$$
$$\operatorname{arcsinh}(r)=\ln(r+\sqrt{r^2+1})$$
where we have the natural exponential function and the natural logarithm.  Particularly, $e\approx2.718281828\dots$  As for logarithms, you'll have to study up on that.
So, we proceed as follows:
$$5=x^3+x$$
Multiply both sides by $\frac{3\sqrt3}2$
$$5\left(\frac{3\sqrt3}2\right)=\left(\frac{3\sqrt3}2\right)(x^3+x)$$
$$\frac{15\sqrt3}2=\frac{3\sqrt3}2x^3+\frac{3\sqrt3}2x$$
Substitute $x=\frac{2\sqrt3}3y$
$$\frac{15\sqrt3}2=\frac{3\sqrt3}2\left(\frac{2\sqrt3}3y\right)^3+\frac{3\sqrt3}2\left(\frac{2\sqrt3}3y\right)$$
$$\frac{15\sqrt3}2=4y^3+3y$$
Use the tigonometric identity
$$\frac{15\sqrt3}2=4y^3+3y=\sinh(3\operatorname{arcsinh}(y))$$
$$\frac{15\sqrt3}2=\sinh(3\operatorname{arcsinh}(y))$$
$$\underbrace{\operatorname{arcsinh}(\sinh}_{\text{cancellation}}(3\operatorname{arcsinh}(y)))=\operatorname{arcsinh}\left(\frac{15\sqrt3}2\right)$$
$$3\operatorname{arcsinh}(y)=\operatorname{arcsinh}\left(\frac{15\sqrt3}2\right)$$
$$\operatorname{arcsinh}(y)=\frac13\operatorname{arcsinh}\left(\frac{15\sqrt3}2\right)$$
$$\underbrace{\sinh(\operatorname{arcsinh}}_{\text{cancellation}}(y)=\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{15\sqrt3}2\right)\right)$$
$$y=\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{15\sqrt3}2\right)\right)$$
Multiply both sides by $\frac{2\sqrt3}3$


$$x=\frac{2\sqrt3}3y=\frac{2\sqrt3}3\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{15\sqrt3}2\right)\right)$$

which is equivalent to Robert's answer.
A: 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 $1$.
A: Another way to write a solution is
$$ x = \dfrac{2}{\sqrt{3}} \sinh\left(\frac{1}{3} \text{arcsinh}\left( \frac{15}{2} \sqrt{3}\right)\right)$$
A: $$  \sqrt[3] { \frac{5}{2} + \sqrt{\frac{679}{108}}   } +   \sqrt[3] { \frac{5}{2} - \sqrt{\frac{679}{108}}   }  \approx 1.51598 $$
OR
$$  \sqrt[3] {  \; \; \sqrt{\frac{679}{108}} + \frac{5}{2} \; \;  } \; \; - \; \;   \sqrt[3] { \; \; \sqrt{\frac{679}{108}} - \frac{5}{2} \; \;   }   \approx 1.51598 $$

A: Sometimes it is difficult to find roots in closed form, so this answer is in the spirit of numerical values for the roots:
$$x^3+x=5$$
$$x^2+1=\frac5x$$
$$x^2=\frac5x-1=\frac{5-x}x$$
$$x=\sqrt{\frac{5-x}x}$$
Once you solve for $x$, you can employ fixed-point iteration.  First, see the root is near $x=1.5$
$$x_0=1.5$$
$$x_1=\sqrt{\frac{5-x_0}{x_0}}=\sqrt{\frac{5-1.5}{1.5}}=\sqrt{\frac{7}{3}}=1.527525231651947$$
$$x_2=\sqrt{\frac{5-{x_1}}{x_1}}=1.507736168412725$$
$$x_3=\sqrt{\frac{5-x_2}{x_2}}=1.521916573662357$$
And you can keep doing this until you find an answer out as many decimals you want.

A much better algorithm to use is Newton's method:
$$x_{n+1}=x_n-\frac{(x_n)^3+x_n-5}{3(x_n)^2+1}$$
Starting with $x_0=1.5$ again,
$$x_0=1.5$$
$$x_1=1.5-\frac{1.5^3+1.5-5}{3(1.5)^2+1}=1.516129032258065$$
$$x_2=1.51598024045$$
Admittedly more complicated, but the amount of digits correct just about doubles each time, which is far faster than the fixed point iteration method.  Also note this already captures all of the digits Will Jagy gives us.
