20. $$ x^{\frac{2}{3}}-5 x^{\frac{1}{3}}+6=0 $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
x^{\frac{2}{3}} - 5x^{\frac{1}{3}} + 6 = 0
$$

**Step 1: Substitute to Simplify**

Let’s make a substitution to simplify the equation. Let:

$$
y = x^{\frac{1}{3}} \quad \Rightarrow \quad y^3 = x
$$

Then:

$$
x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2 = y^2
$$

Substituting into the original equation:

$$
y^2 - 5y + 6 = 0
$$

**Step 2: Solve the Quadratic Equation**

The equation $$ y^2 - 5y + 6 = 0 $$ is a standard quadratic equation. We can factor it:

$$
y^2 - 5y + 6 = (y - 2)(y - 3) = 0
$$

So, the solutions for $$ y $$ are:

$$
y = 2 \quad \text{or} \quad y = 3
$$

**Step 3: Back-Substitute to Find $$ x $$**

Recall that $$ y = x^{\frac{1}{3}} $$. Therefore:

1. If $$ y = 2 $$:

$$
x^{\frac{1}{3}} = 2 \quad \Rightarrow \quad x = 2^3 = 8
$$

2. If $$ y = 3 $$:

$$
x^{\frac{1}{3}} = 3 \quad \Rightarrow \quad x = 3^3 = 27
$$

**Step 4: Verify the Solutions**

It's always good practice to verify the solutions in the original equation:

1. For $$ x = 8 $$:

$$
8^{\frac{2}{3}} - 5 \times 8^{\frac{1}{3}} + 6 = (4) - (5 \times 2) + 6 = 4 - 10 + 6 = 0
$$

2. For $$ x = 27 $$:

$$
27^{\frac{2}{3}} - 5 \times 27^{\frac{1}{3}} + 6 = (9) - (5 \times 3) + 6 = 9 - 15 + 6 = 0
$$

Both solutions satisfy the original equation.

**Final Answer:**

The solutions are $$ x = 8 $$ and $$ x = 27 $$.
The solutions are $$ x = 8 $$ and $$ x = 27 $$.

20. \( x^{\frac{2}{3}}-5 x^{\frac{1}{3}}+6=0 \)