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This ones a good one

mmm unknown the denominator , must multiply both sides by a square!

so we get (9x^2 - 64 ) ( x^2 -25)^2 > 0
now lets break it down using difference of sqaures

(3x-8)(3x+8) [ (x-5)(x+5)]^2 >0
now distribute the square on the last terms ( so to speak )

What would that solution look like in interval notation?

somebody dig us out of this hole!! lol

I figured out where I noobed up

when I multiplied top and bottom by the sqaure , I didnt cancel a factor on the LHS

Ah so, now x=0 is a possibility.

what happened to the denominator?

Going to take a shot.
The fraction has to be greater than 0.
Hence
x>0 and not = to 5
Refer to the plot attachment

*numerator and denominator can never be of opposite signs

OK. It's got me to thinking lol.

\[\left(x>-\frac{8}{3} \text{and } x<\frac{8}{3}\right)\text{or } (x<-5) \text{ or }( x>5) \]