Answers: 1
is real, then 
will always be non-negative, so the sign of the numerator will determine the solution to the inequality. In order for the expression on the left hand side to be defined, we require![(x^3-7)^2\neq0\implies x^3\neq7\implies x\neq\sqrt[3]7\approx1.91](/tpl/images/0021/6199/03965.png)
, say 
. This gives the numerator a value of 
.
, we can pick 
and we get 
.
, we can pick 
and get 
. But remember, we can't let ![x=\sqrt[3]7](/tpl/images/0021/6199/331cd.png)
.![\left(-\infty,-\dfrac35\right]\cup\left(\dfrac23,\sqrt[3]7\right)\cup\left(\sqrt[3]7,\infty\right)](/tpl/images/0021/6199/6a74d.png)
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Solve the inequality/solve for x: [tex]\frac{(5x+3)(3x-2)}{(x^3-7)^2} \geq 0[/tex]...
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