# Solving Polynomial Inequalities of x3<x ?

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### 1 Answer

The inequality to be solved is : x^3 < x

x^3 < x

move all the terms to the left

=> x^3 - x < 0

factor the left hand side

=> x(x^2 - 1)< 0

=> x(x - 1)(x + 1) < 0

Now the left hand side is less than 0 if

- all the factors are negative

=> x < 0 , x - 1 < 0 and x + 1 < 0

=> x < 0 , x < 1 and x < -1

x < -1 satisfies all the conditions

- only one of the factors is less than 0

1. x < 0, x - 1 > 0 and x + 1> 0

=> x < 0 , x > 1 and x > -1

no value of x can satisfy this

2. x > 0 , x + 1 < 0 and x - 1 > 0

=> x > 0 , x < -1 and x > 1

no value of x can satisfy this

3. x > 0, x + 1 > 0 and x - 1 < 0

=> x > 0, x > -1 and x < 1

this is satisfied by x > 0 and x < 1

**So the values of x that satisfy x^3 < x lie in (-inf., -1)U(0 , 1)**