How do you factor #125e^3-27f^3#?

2 Answers
Jul 30, 2018

Answer:

#(5e-3f)(25e^2+15ef+9f^2)#

Explanation:

#"this is a "color(blue)"difference of cubes"#

#•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)#

#125e^3=(5e)^3rArra=5e#

#27f^3=(3f)^3rArrb=3f#

#=(5e-3f)((5e)^2+5e*3f+(3f)^2)#

#=(5e-3f)(25e^2+15ef+9f^2)#

Jul 30, 2018

Answer:

#(5e-3f)(25e^2+15ef+9f^2)#

Explanation:

What we have is a difference of cubes, which factors as follows:

#bar(ul|color(white)(2/2)(a^3-b^3)=(a-b)(a^2+ab+b^2)color(white)(2/2)|#

We have the following:

#125e^3-27f^3#, where #a=root3(125e^3)#, and #b=root3(27f^3)#.

Simplifying these, we get #a=5e# and #b=3f#. Let's plug these into our difference of cubes expression to get

#(5e-3f)(25e^2+15ef+9f^2)#

Hope this helps!