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

Question

1812 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-24T11:23:47

See a solution process below:

This is a special case of a quadratic:

$a^{2} - b^{2} = (a + b) (a - b)$ $color(red)(a)^2 - color(blue)(b)^2 = (color(red)(a) + color(blue)(b))(color(red)(a) - color(blue)(b))$

Let $a = 3 x^{2} y$ $color(red)(a) = 3x^2y$ ; Then $a^{2} = 9 x^{4} y^{2}$ $color(red)(a)^2 = 9x^4y^2$
Let $b = z^{3}$ $color(blue)(b) = z^3$ ; Then $b^{2} = z^{6}$ $color(blue)(b)^2 = z^6$

Substituting gives:

$9 x^{4} y^{2} - z^{6} \Rightarrow$ $color(red)(9x^4y^2) - color(blue)(z^6) =>$

${(3 x^{2} y)}^{2} - {(z^{3})}^{2} \Rightarrow$ $color(red)((3x^2y))^2 - color(blue)((z^3))^2 =>$

$(3 x^{2} y + z^{3}) (3 x^{2} y - z^{3})$ $(color(red)(3x^2y) + color(blue)(z^3))(color(red)(3x^2y) - color(blue)(z^3))$

How do you factor 9x^4y^2 - z^69x4y2−z69x^4y^2 - z^6?