How do you factor $x^6 - 27y^3 - x^4 - 3x^2y - 9y^2$ ?

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Answer 1 · 2015-06-01T18:09:22

$x^6-27y^3-x^4-3x^2y-9^2$

$=(x^2)^3-(x^2)^2-(x^2)(3y)-(3y)^2-(3y)^3$

$= X^3-X^2-XY-Y^2-Y^3$

...where $X=x^2$ and $Y=3y$ .

$= (X^3-Y^3)-(X^2+XY+Y^2)$

$= (X-Y)(X^2+XY+Y^2) - (X^2+XY+Y^2)$

$= (X-Y-1)(X^2+XY+Y^2)$

$= ((x^2)-(3y)-1)((x^2)^2+(x^2)(3y)+(3y)^2)$

$= (x^2-3y-1)(x^4+3x^2y+9y^2)$

How do you factor x^6 - 27y^3 - x^4 - 3x^2y - 9y^2x^6 - 27y^3 - x^4 - 3x^2y - 9y^2?