# Solve for x From the Inequation

## Algebra Math Problem: Solve for x when cube root of 3x – 2 is less than x

When x is a real number, Solve for x from the inequation “cube root of 3x – 2 is less than x”.

\sqrt[3]{3x-2}< x

## Solving the inequation

\sqrt[3]{3x-2}< x

Take cube in both sides then

3x − 2 < x³

Rearrange this equation then

x³ − 3x + 2 > 0

### Factorise the cubic inequation x³ – 3x + 3 > 0, then

x³ – 3x + 2 = x³ – x – 2x + 2

⇒ x³ – 3x + 2 = x(x² – 1) – 2(x – 1)

We know that x² – 1 = (x – 1)(x+1)

⇒ x³ – 3x + 2 = x(x – 1)(x+1) – 2(x – 1)

⇒ x³ – 3x + 2 = (x – 1)(x(x+1) – 2)

So, x³ – 3x + 2 = (x – 1)(x² + x – 2)

### Factorise x² + x – 2

⇒ x³ – 3x + 2 = (x – 1)(x² – x + 2x – 2)

⇒ x³ – 3x + 2 = (x – 1)(x(x – 1) + 2(x – 1))

So, x³ – 3x + 2 = (x – 1)(x – 1)(x + 2)

⇒ x³ – 3x + 2 = (x – 1)² (x + 2)

So we can write

(x – 1)² (x + 2) > 0

Where (x – 1)² is greater than 0 (because squares are always positive for real numbers)

Then, (x + 2) > 0

⇒ x > -2

So the solution to the math problem is **x > -2**