If [tex]x[/tex] is real, then [tex](x^3-7)^2[/tex] 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
[tex](x^3-7)^2\neq0\implies x^3\neq7\implies x\neq\sqrt[3]7\approx1.91[/tex]
We have
[tex](5x+3)(3x-2)=0\implies x=-\dfrac35=-0.6,x=\dfrac23\approx0.66[/tex]
so there are three intervals we need to check.
(1) We pick a value of [tex]x[/tex] from [tex]-\infty<x<-\dfrac35[/tex], say [tex]x=-1[/tex]. This gives the numerator a value of [tex](-5+3)(-3-2)=(-2)(-5)=10>0[/tex].
(2) From [tex]-\dfrac35<x<\dfrac23[/tex], we can pick [tex]x=0[/tex] and we get [tex](0+3)(0-2)=-6<0[/tex].
(3) From [tex]\dfrac23<x<\infty[/tex], we can pick [tex]x=1[/tex] and get [tex](5+3)(3-2)=8>0[/tex]. But remember, we can't let [tex]x=\sqrt[3]7[/tex].
So the solution to the inequality is the union of the two intervals that we showed would make the numerator positive, while still avoiding making the expression undefined:
[tex]\left(-\infty,-\dfrac35\right]\cup\left(\dfrac23,\sqrt[3]7\right)\cup\left(\sqrt[3]7,\infty\right)[/tex]
