Inequality

Method 1:

Think of the number line and the main 4 regions:
#a) \(x < -1\)
#b) \(-1 < x < 0\)
#c) \(0 < x < 1\)
#d) \(x > 1\)

Try a number from each of these regions to check the inequality \(x > x^3\)
For #a) x = -2: -2 > -8 --- satisfies
For #b) x = -1/2: -1/2 > -1/8 --- does not satisfy
For #c) x = 1/2: 1/2 > 1/8 --- satisfies
For #d) x = 2: 2 > 8 --- does not satisfy

Thus, x can be in regions #a or #c
=> \(x < -1\) or \(0 < x < 1\)


Method 2:

\(x > x^3\\
=> x(x^2 - 1) < 0\\
=> x(x + 1)(x - 1) < 0\)

Trying x = 2, we have: x(x + 1)(x - 1) = 6 > 0
Thus, we have the curve of x(x + 1)(x - 1) as shown below. Since we needed to solve for \(x(x + 1)(x - 1) < 0\), the required solutions (regions) which are negative, i.e. below the X-axis, are shown in yellow:



Method 3:

\(x > x^3\\
=> x(x^2 - 1) < 0\\
=> x(x + 1)(x - 1) < 0\)

We need:

2 of the above 3 terms positive and 1 negative; i.e. x and (x + 1) are positive and (x - 1) is negative => \(0 < x < 1\)

Or

All 3 of the above terms negative: i.e. (x -1), x and (x + 1) are negative => x + 1 < 0 => \(x < -1\)