$$ x ^ { \frac { 4 } { 3 } } - \sqrt[ 3 ] { x ^ { 2 } } - 72 = 0 $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
x^{\frac{4}{3}} - \sqrt[3]{x^2} - 72 = 0
$$

**Step 1: Substitute**

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

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

Then:

$$
x = y^3
$$

Now, express the original equation in terms of $$ y $$:

$$
(x^{\frac{4}{3}}) = (y^3)^{\frac{4}{3}} = y^4
$$
$$
(\sqrt[3]{x^2}) = (y^3)^{\frac{2}{3}} = y^2
$$

Substituting these into the equation:

$$
y^4 - y^2 - 72 = 0
$$

**Step 2: Solve the Quartic Equation**

Let’s set $$ t = y^2 $$. The equation becomes:

$$
t^2 - t - 72 = 0
$$

Solve for $$ t $$ using the quadratic formula:

$$
t = \frac{1 \pm \sqrt{1 + 288}}{2} = \frac{1 \pm \sqrt{289}}{2} = \frac{1 \pm 17}{2}
$$

This gives:

$$
t = 9 \quad \text{or} \quad t = -8
$$

Since $$ t = y^2 $$ and $$ y^2 $$ cannot be negative:

$$
t = 9 \implies y^2 = 9 \implies y = \pm 3
$$

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

Recall that:

$$
y = x^{\frac{1}{3}} \implies x = y^3
$$

Thus:

1. If $$ y = 3 $$:

$$
x = 3^3 = 27
$$

2. If $$ y = -3 $$:

$$
x = (-3)^3 = -27
$$

**Step 4: Verify Solutions**

Plugging back into the original equation:

- For $$ x = 27 $$:
  
  $$
  27^{\frac{4}{3}} - \sqrt[3]{27^2} - 72 = 81 - 9 - 72 = 0
  $$

- For $$ x = -27 $$:
  
  $$
  (-27)^{\frac{4}{3}} - \sqrt[3]{(-27)^2} - 72 = 81 - 9 - 72 = 0
  $$

Both solutions satisfy the original equation.

**Final Answer:**

The real solutions are $$ x = 27 $$ and $$ x = -27 $$.
The solutions are $$ x = 27 $$ and $$ x = -27 $$.

\( x ^ { \frac { 4 } { 3 } } - \sqrt[ 3 ] { x ^ { 2 } } - 72 = 0 \)