Answer
$(-\infty, -1) \cup (1,7) $
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$x(1-x^2)^3 \gt 7(1-x^2)^3$,
$x(1-x^2)^3-7(1-x^2)^3 \gt 0$,
$(x-7)(1-x^2)^3 \gt 0$,
$(x-7)(1-x^6) \gt 0$,
$(x-7)(1-x^3)(1+x^3) \gt 0$,
$(x-7)(1-x)(x^2+x+1)(1+x^3) \gt 0$
$f(x)=(x-7)(1-x)(x^2+x+1)(1+x^3)$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(x-7)(1-x)(x^2+x+1)(1+x^3)=0$
$x=1$ or $x=7$ or $x=-1$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \gt 0 ? \\
& &(a-7)(1-a)(a^2+a+1)(1+a^3)& \\
(-\infty, -1) & -2 & (-)(+)(+)(-) & T\\
(-1, 1) & 0 & (-)(+)(+)(+) & F\\
(1, 7) & 5 & (-)(-)(+)(+) &T\\
(7,\infty) & 9 & (+)(-)(+)(+) & F
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\infty, -1) \cup (1,7) $
