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

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.
