91Ó°ÊÓ

In dealing with radical equations, one of the first steps can be to simplify the equation by introducing a new variable. This makes the equation easier to handle. For example, if we start with the equation \[ x^{2/3} - x^{1/3} - 20 = 0 \], it can be simplified by letting \[ y = x^{1/3} \]. This substitution transforms our original problem into a more familiar form. \
By making the substitution, \[ y^2 = x^{2/3} \], the equation becomes \[ y^2 - y - 20 = 0 \]. Now, instead of working with radicals, we've transformed the problem into solving a quadratic equation—something much more straightforward!

Now that we have simplified our radical equation, we are left with a quadratic equation: \[ y^2 - y - 20 = 0 \]. Quadratic equations take the form of \[ ax^2 + bx + c = 0 \]. In this case, \[ a = 1 \], \[ b = -1 \], and \[ c = -20 \].
To solve this quadratic equation, we can factor it: \[ y^2 - y - 20 = 0 \].
This factors into \[ (y - 5)(y + 4) = 0 \], giving us the solutions \[ y = 5 \] and \[ y = -4 \]. These are the solutions for \[ y \], but remember, our original variable was \[ x \]. So, the next step is to substitute \[ y \] back into our expression involving \[ x \].

The final and very important step is to verify that our solutions are correct by substituting them back into the original equation. Let's consider both solutions:
For \[ y = 5 \]: We had \[ x^{1/3} = 5 \], hence \[ x = 5^3 = 125 \]. Substituting this into the original equation \[ x^{2/3} - x^{1/3} - 20 = 0 \], we get: \[ 125^{2/3} - 125^{1/3} - 20 \]. Simplified, this becomes: \[ 25 - 5 - 20 = 0 \] which holds true.
For \[ y = -4 \]: We had \[ x^{1/3} = -4 \], hence \[ x = (-4)^3 = -64 \]. Substituting this into the original equation, we get: \[ (-64)^{2/3} - (-64)^{1/3} - 20 = 0 \]. Simplified, this becomes: \[ 16 - (-4) - 20 = 0 \] which also holds true.
Thus, both \[ x=125 \] and \[ x=-64 \] are verified as real solutions to the original radical equation.