How do you factor $8 x^{3} y^{6} - 125$ $8x^3y^6-125$ ?

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How do you factor $8 x^{3} y^{6} - 125$ $8x^3y^6-125$ ?

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Answer 1 · 2016-01-03T06:54:10

$(2 x y^{2} - 5) (2 x^{2} y^{4} + 10 x y^{2} + 25)$ $(2xy^2-5)(2x^2y^4+10xy^2+25)$

Explanation:

This is a difference of cubes, which takes the general form:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

Recognize that $8 x^{3} y^{6} = {(2 x y^{2})}^{3}$ $8x^3y^6=(2xy^2)^3$ and that $125 = 5^{3}$ $125=5^3$ .

${(2 x y^{2})}^{3} - 5^{3} = (2 x y^{2} - 5) ({(2 x y^{2})}^{2} + 2 x y^{2} (5) + 5^{2})$ $(2xy^2)^3-5^3=(2xy^2-5)((2xy^2)^2+2xy^2(5)+5^2)$

$= (2 x y^{2} - 5) (2 x^{2} y^{4} + 10 x y^{2} + 25)$ $=(2xy^2-5)(2x^2y^4+10xy^2+25)$

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How do you factor $8 x^{3} y^{6} - 125$ $8x^3y^6-125$ ?

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