To solve the equation [tex]\( x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0 \)[/tex], we can use the hint and make a substitution to transform the equation into a simpler form. Let's proceed step by step.
### Step 1: Substitution
Let [tex]\( u = x^{\frac{1}{3}} \)[/tex]. This means [tex]\( u^3 = x \)[/tex].
Substitute [tex]\( u \)[/tex] into the original equation:
[tex]\[ x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0 \][/tex]
In terms of [tex]\( u \)[/tex]:
[tex]\[ (u^3)^{\frac{2}{3}} - (u^3)^{\frac{1}{3}} - 6 = 0 \][/tex]
[tex]\[ u^2 - u - 6 = 0 \][/tex]
### Step 2: Solve the Quadratic Equation
Now we solve the quadratic equation [tex]\( u^2 - u - 6 = 0 \)[/tex].
The quadratic formula is given by:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( u^2 - u - 6 = 0 \)[/tex], we have [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \)[/tex]. Plug these values into the quadratic formula:
[tex]\[ u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{1 \pm \sqrt{1 + 24}}{2} \][/tex]
[tex]\[ u = \frac{1 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ u = \frac{1 \pm 5}{2} \][/tex]
This gives us two solutions:
[tex]\[ u = \frac{1 + 5}{2} = 3 \][/tex]
[tex]\[ u = \frac{1 - 5}{2} = -2 \][/tex]
### Step 3: Back-substitute [tex]\( u \)[/tex] to find [tex]\( x \)[/tex]
Recall that [tex]\( u = x^{\frac{1}{3}} \)[/tex]. Now we back-substitute to find [tex]\( x \)[/tex].
1. For [tex]\( u = 3 \)[/tex]:
[tex]\[ 3 = x^{\frac{1}{3}} \][/tex]
Cube both sides:
[tex]\[ 3^3 = x \][/tex]
[tex]\[ x = 27 \][/tex]
2. For [tex]\( u = -2 \)[/tex]:
[tex]\[ -2 = x^{\frac{1}{3}} \][/tex]
Cube both sides:
[tex]\[ (-2)^3 = x \][/tex]
[tex]\[ x = -8 \][/tex]
### Step 4: Final Answer
The solutions to the equation [tex]\( x^{\frac{2}{3}} - x^{\frac{1}{3}} - 6 = 0 \)[/tex] are:
[tex]\[ x = 27 \][/tex]
[tex]\[ x = -8 \][/tex]
So, the exact solutions are:
[tex]\[ \boxed{-8 \quad \text{and} \quad 27} \][/tex]
