MHB Factoring Fractional Expression

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The expression (x^3/8) - (512/x^3) can be simplified by finding a common denominator, resulting in (x^6 - 4096)/8x^3. The numerator can be factored as [(x^3 - 64)(x^3 + 64)], recognizing that x^6 - 4096 is a difference of squares. Further factoring reveals both a sum and difference of cubes in the numerator. The discussion emphasizes the need to explore additional factorization options for a complete solution. The final answer is not yet reached, indicating further simplification is possible.
mathdad
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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?
 
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Yes, in fact:

$$x^6-4096=x^6-2^{12}=\left(x^3\right)^2-\left(2^6\right)^2=\left(x^3+2^6\right)\left(x^3-2^6\right)=\left(x^3+\left(2^2\right)^3\right)\left(x^3-\left(2^2\right)^3\right)$$
 
Is that the final answer? Can we reduce it further?
 
RTCNTC said:
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
 
I will complete the problem when time allows.
 
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