Question 179673
You can temporarily chang variables to facilitate solving problems like this:
{{{(x-5)^(2/3) + 3(x-5)^(1/3)-4 = 0}}} You can rearrange this as:
{{{((x-5)^(1/3))^2+3((x-5)^(1/3))-4 = 0}}}
 Let's substitute {{{y = (x-5)^(1/3)}}} to give us:
{{{y^2+3y-4 = 0}}} Factor this:
{{{(y-1)(y+4) = 0}}} from which:
{{{y = 1}}} or {{{y = -4}}} Replacing the variable: {{{y = (x-5)^(1/3)}}} we get:
{{{(x-5)^(1/3) = 1}}} or {{{(x-5)^(1/3) = -4}}} so...
If {{{(x-5)^(1/3) = 1}}} Raise each side to the third power.
{{{x-5 = 1^3}}} but {{{1^3 = 1}}} so...
{{{x-5 = 1}}} Add 5 to both sides.
{{{highlight(x = 6)}}} and...
{{{(x-5)^(1/3) = -4}}} Cube both sides.
{{{x-5 = (-4)^3}}}
{{{x-5 = -64}}} Add 5 to both sides.
{{{highlight(x = -59)}}}