Question 230532
The key to the simple solution for this equation is to recognize that {{{x^(2/3) = (x^(1/3))^2}}}. Once we see this, we can recognize that your equation is a quadratic equation in {{{x^(1/3)}}} and that this quadratic equation will factor fairly easily:
{{{(x^(1/3) - 5)(x^(1/3) + 4) = 0}}}
(If, after thinking about this, you still have trouble seeing this, see below for a way to solve this that is longer but may be clearer.)
By the Zero Product Property we know that this product is zero only if one of the factors is zero. So:
{{{x^(1/3) - 5 = 0}}} or {{{x^(1/3) + 4 = 0}}}
Solving these we get:
{{{x^(1/3) = 5}}} or {{{x^(1/3) = -4}}}
Cubing both sides:
{{{x = 125}}} or {{{x = -64}}}<br>
A longer, perhaps clearer, solution. We still will be using {{{x^(2/3) = x^(1/3)^2}}}. But this time we will use a temporary variable to make things clearer.
Let {{{q = x^(1/3)}}}
then {{{q^2 = (x^(1/3))^2 = x^(2/3)}}}
Now we'll substitute these into your equation:
{{{q^2 - q - 20 = 0}}}
This is clearly a quadratic equation and the factoring is more obvious:
{{{(q-5)(q+4)=0}}}
Solving this we get:
{{{q-5 = 0}}} or {{{q+4=0}}}
{{{q = 5}}} or {{{q = -4}}}
At this point we have solved for q. But we want to solve for x. So we substitute back in for q:
{{{x^(1/3) = 5}}} or {{{x^(1/3) = -4}}}
To solve for x we cube both sides of this:
{{{x = 125}}} or {{{x = -64}}}<br>
If you have to use the temporary variables to solve problems like this, it would be good, after a while, to wean yourself off the temporary variables.