Question

X^3>x^2 and |x| > 1, then
1. X > 1

2. X < -1

3. -1 < X < 1

4. X < -1 or X > 1

5. X < -1 and X > 1

Answer 1

Answer:

1. X  > 1

Step-by-step explanation:

We know $x^2 > 1$ because $x^2 = |x|^2 > 1$. That's why

$X^3 > x^2 > 1$.

We can now subtract 1 in both sides of the inequality:

$X^3 -1 > 0$.

Factoring $X^3 -1$ as a difference of cubes, we get:

$(X-1)(X^2 + X + 1) > 0$.

Thus, we have two factors whose multiplication is positive. Then, both are positive or both are negative. The second case is impossible, because $X^2 + X + 1$ can never be positive. The reason is the following:

$X^2 + X + 1 = 2 \frac{X^2 + X + 1}{2} = \frac{2X^2 + 2X + 2}{2} = \frac{X^2 + 2X + 1 +X^2 + 1}{2} = \frac{(X+1)^2 +X^2 + 1}{2}$

witch is always positive because $(X+1)^2 + X^2 + 1$ is a sum of squares, and the squares are always positive.

We conclude that both $(X-1)$ and $(X^2 + X + 1)$ have to be positive, and then $X-1 > 0$ implies $X > 1$