Originally Posted by

**BobP** At the point where you have $\displaystyle 8x+3 = 64 + .....$ , rearrange the equation as

$\displaystyle 7(x-4) - 12(x-4)^{2/3}-48(x-4)^{1/3} - 29 = 0.$

Now make a substitution,$\displaystyle z = (x-4)^{1/3}, $ and you arrive at the cubic equation

$\displaystyle 7z^{3}-12z^{2}-48z-29 = 0.$

Whether or not you accept trial values for $\displaystyle z$ (which is a normal method) I don't know. ($\displaystyle z = -1$ is a fairly obvious solution.) The alternative is to use perhaps Cardano's method.