Solve the inequality (x^3 + 1)(x^2 - 2) < x(x^3 + 1).

Step-by-step solution

1. 1

   Move everything to one side — never divide by $x^{3}+1$. Dividing by an expression of unknown sign would silently flip the inequality on part of the line. Instead subtract:

   $\left(x^{3}+1\right)\left(x^{2}-2\right)-x\left(x^{3}+1\right)<0$

   and factor out the shared $x^{3}+1$:

   $\left(x^{3}+1\right)\left(x^{2}-x-2\right)<0$

2. 2

   Factor both pieces completely. As a sum of cubes and a simple trinomial:

   $x^{3}+1=(x+1)\left(x^{2}-x+1\right),\qquad x^{2}-x-2=(x-2)(x+1)$

   so the inequality becomes

   $(x+1)^{2}\left(x^{2}-x+1\right)(x-2)<0$

   The factor $(x+1)$ appears twice, which is the key structural fact.

3. 3

   Eliminate the factors that never change sign. $(x+1)^{2}\ge 0$ always, with equality only at $x=-1$. And $x^{2}-x+1$ has discriminant $(-1)^{2}-4=-3<0$ with positive leading coefficient, so it is strictly positive for every real $x$. Neither can make the product negative.

4. 4

   Reduce to a single sign condition. With the two non-negative factors set aside, the product is negative exactly when $x-2<0$ and neither non-negative factor is zero:

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

   At $x=-1$ the whole product is $0$, and $0<0$ is false, so $-1$ must be punched out.

5. 5

   Write the solution set.

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

   $x=2$ is excluded too, since the inequality is strict.

6. 6

   Spot-check the intervals. Let $g(x)$ be the left side minus the right side. Then $g(-3)=-260<0$ and $g(0)=-2<0$ (both inside the solution set), while $g(-1)=0$ and $g(2)=0$ (excluded) and $g(3)=112>0$ (outside). This also shows the graph touches zero at $x=-1$ without crossing — the signature of an even-multiplicity root.