How do you factor $27 w^{3} z - z^{4} w^{6}$ $27w^3z-z^4w^6$ ?

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How do you factor $27 w^{3} z - z^{4} w^{6}$ $27w^3z-z^4w^6$ ?

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Answer 1 · 2017-04-07T06:15:13

$27 w^{3} z - z^{4} w^{6} = w^{3} z (3 - z w) (9 + 3 z w + z^{2} w^{2})$ $27w^3z-z^4w^6 = w^3z(3-zw)(9+3zw+z^2w^2)$

Explanation:

The difference of cubes identity can be written:

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

We will use this with $a = 3$ $a=3$ and $b = z w$ $b=zw$ .

$color(white)()$
Given:

$27 w^{3} z - z^{4} w^{6}$ $27w^3z-z^4w^6$

First note that both of the terms are divisible by $w^{3}$ $w^3$ and by $z$ $z$ , so by $w^{3} z$ $w^3z$ . So we can separate that out as a factor first:

$27 w^{3} z - z^{4} w^{6} = w^{3} z (27 - z^{3} w^{3})$ $27w^3z-z^4w^6 = w^3z(27-z^3w^3)$

$27 w^{3} z - z^{4} w^{6} = w^{3} z (3^{3} - {(z w)}^{3})$ $color(white)(27w^3z-z^4w^6) = w^3z(3^3-(zw)^3)$

$27 w^{3} z - z^{4} w^{6} = w^{3} z (3 - z w) (3^{2} + 3 z w + {(z w)}^{2})$ $color(white)(27w^3z-z^4w^6) = w^3z(3-zw)(3^2+3zw+(zw)^2)$

$27 w^{3} z - z^{4} w^{6} = w^{3} z (3 - z w) (9 + 3 z w + z^{2} w^{2})$ $color(white)(27w^3z-z^4w^6) = w^3z(3-zw)(9+3zw+z^2w^2)$

This is as far as we can go with Real coefficients.

The remaining quartic factor $(9 + 3 z w + z^{2} w^{2})$ $(9+3zw+z^2w^2)$ can be factored further, but only with the use of Complex coefficients.

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How do you factor $27 w^{3} z - z^{4} w^{6}$ $27w^3z-z^4w^6$ ?

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How do you factor $27 w^{3} z - z^{4} w^{6}$ $27w^3z-z^4w^6$ ?