Factoring Fractional Expression

[mathdad]

Factor (x^3/8) - (512/x^3).

LCD = 8x^3

(x^6 - 8(512))/8x^3

(x^6 - 4096)/8x^3

[(x^3 - 64)(x^3 + 64)]/8x^3

In the numerator, the expression (x^3 - 64) is the difference of cubes, right?

 

[MarkFL]

Yes, in fact:

\(\displaystyle x^6-4096=x^6-2^{12}=\left(x^3\right)^2-\left(2^6\right)^2=\left(x^3+2^6\right)\left(x^3-2^6\right)=\left(x^3+\left(2^2\right)^3\right)\left(x^3-\left(2^2\right)^3\right)\)

 

[mathdad]

Is that the final answer? Can we reduce it further?

 

[MarkFL]

Is that the final answer? Can we reduce it further?

It can be factored further, as there is both a sum and difference of cubes there. Bear in mind I only dealt with the numerator. :D

 

[mathdad]

I will complete the problem when time allows.