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factor ( 512x^{3}y^{6} - 125 ) completely.

Question

factor ( 512x^{3}y^{6} - 125 ) completely.

Explanation:

Step1: Identify the difference of cubes

We know that \(a^3 - b^3=(a - b)(a^2+ab + b^2)\). Here, \(512x^{3}y^{6}=(8xy^{2})^{3}\) and \(125 = 5^{3}\). So we can rewrite the expression as a difference of cubes: \((8xy^{2})^{3}-5^{3}\).

Step2: Apply the difference of cubes formula

Using the formula \(a^3 - b^3=(a - b)(a^2+ab + b^2)\) with \(a = 8xy^{2}\) and \(b = 5\), we get:
First, \(a - b=8xy^{2}-5\).
Then, \(a^{2}=(8xy^{2})^{2}=64x^{2}y^{4}\), \(ab=(8xy^{2})\times5 = 40xy^{2}\), and \(b^{2}=5^{2}=25\). So \(a^{2}+ab + b^{2}=64x^{2}y^{4}+40xy^{2}+25\).
Putting it together, the factored form is \((8xy^{2}-5)(64x^{2}y^{4}+40xy^{2}+25)\).

Answer:

\((8xy^{2}-5)(64x^{2}y^{4}+40xy^{2}+25)\)