How do you factor #9x^4y^2 - z^6#?

1 Answer
Mar 24, 2018

Answer:

See a solution process below:

Explanation:

This is a special case of a quadratic:

#color(red)(a)^2 - color(blue)(b)^2 = (color(red)(a) + color(blue)(b))(color(red)(a) - color(blue)(b))#

  • Let #color(red)(a) = 3x^2y#; Then #color(red)(a)^2 = 9x^4y^2#

  • Let #color(blue)(b) = z^3#; Then #color(blue)(b)^2 = z^6#

Substituting gives:

#color(red)(9x^4y^2) - color(blue)(z^6) =>#

#color(red)((3x^2y))^2 - color(blue)((z^3))^2 =>#

#(color(red)(3x^2y) + color(blue)(z^3))(color(red)(3x^2y) - color(blue)(z^3))#