How do you factor completely $27 x^{3} - 125 y^{3}$ $27x^3-125y^3$ ?

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How do you factor completely $27 x^{3} - 125 y^{3}$ $27x^3-125y^3$ ?

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Answer 1 · 2016-04-26T21:30:50

$27 x^{3} - 125 y^{3} = (3 x - 5 y) (9 x^{2} + 15 x y + 25 y^{2})$ $27x^3-125y^3=(3x-5y)(9x^2+15xy+25y^2)$

Explanation:

Notice that both of the terms are perfect cubes:

$27 x^{3} = {(3 x)}^{3}$ $27x^3 = (3x)^3$

$125 y^{3} = {(5 y)}^{3}$ $125y^3 = (5y)^3$

So we can conveniently use the difference of cubes identity:

$A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$ $A^3-B^3=(A-B)(A^2+AB+B^2)$

with $A = 3 x$ $A=3x$ and $B = 5 y$ $B=5y$ as follows:

$27 x^{3} - 125 y^{3}$ $27x^3-125y^3$

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

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

$= (3 x - 5 y) (9 x^{2} + 15 x y + 25 y^{2})$ $=(3x-5y)(9x^2+15xy+25y^2)$

The remaining quadratic factor has no linear factors with Real coefficients. If we allow Complex coefficients then we can factor a little further:

$= (3 x - 5 y) (3 x - 5 ω y) (3 x - 5 ω^{2} y)$ $=(3x-5y)(3x-5omega y)(3x-5 omega^2 y)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ $1$ .

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How do you factor completely $27 x^{3} - 125 y^{3}$ $27x^3-125y^3$ ?

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