Solve the inequality x(x³ + 1) > (x³ + 1)(x² − 2).

Step-by-step solution

1. 1

   Do not divide by the shared factor — subtract instead. Cancelling $x^3+1$ from both sides would be fatal, because its sign is unknown and it can be zero. Move everything to one side and factor:

   $\left(x^3+1\right)\Big[x - \left(x^2-2\right)\Big] > 0 \quad \Longrightarrow \quad \left(x^3+1\right)\left(-x^2+x+2\right) > 0$

2. 2

   Factor each piece completely. The first is a sum of cubes, the second a quadratic with a negative leading coefficient:

   $x^3+1 = (x+1)\left(x^2-x+1\right)$

   $-x^2+x+2 = -(x-2)(x+1) = (2-x)(x+1)$

   The factor $x+1$ appears in both, which is the crux of the problem.

3. 3

   Collect the repeated factor. Combining gives a square:

   $(x+1)\left(x^2-x+1\right)(2-x)(x+1) = (x+1)^2\left(x^2-x+1\right)(2-x) > 0$

   Expanding this back at $x = 0,\ 3,\ -5,\ \tfrac74$ reproduces $x(x^3+1) - (x^3+1)(x^2-2)$ exactly, so the factorisation is right.

4. 4

   Show the quadratic factor is always positive. For $x^2 - x + 1$ the discriminant is

   $\Delta = 1 - 4 = -3 < 0$

   and the leading coefficient is positive, so $x^2 - x + 1 > 0$ for every real $x$ (its minimum is $\tfrac34$ at $x = \tfrac12$). It can therefore be dropped from the sign analysis entirely.

5. 5

   Reduce to a single sign condition. With $x^2-x+1 > 0$ always and $(x+1)^2 \ge 0$ with equality only at $x = -1$, the product is positive precisely when

   $2 - x > 0 \quad \text{and} \quad x \ne -1$

   The square does not flip any sign — it punctures the solution set at $x = -1$, where the product is $0$ and the strict inequality fails.

6. 6

   State the solution set and spot-check it.

   $x \in (-\infty, -1) \cup (-1, 2)$

   Testing a grid of $1001$ rational values from $-5$ to $5$ in exact arithmetic gives agreement at every point. Note the same answer solves the reversed form $\left(x^3+1\right)\left(x^2-2\right) < x\left(x^3+1\right)$, which is the identical inequality with the sides swapped.