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

Question

1681 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-01T18:09:22

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

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

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

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

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

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

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

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

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

How do you factor x^6 - 27y^3 - x^4 - 3x^2y - 9y^2x6−27y3−x4−3x2y−9y2x^6 - 27y^3 - x^4 - 3x^2y - 9y^2?