How do you factor #125x^3+144#?

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Answer 1 · 2015-12-13T00:51:37

#125x^3 + 144 =(5x + 144^(1/3))(25x^2 - 5(144^(1/3))x + 144^(2/3))#

The sum of two cubes can be factored as
#a^3 + b^3 = (a+b)(a^2 -ab + b^2)#

While unfortunately 144 is not a perfect cube, it still a cube of some number. Specifically it is the cube of #144^(1/3)#

Applying this, we have

#125x^3 + 144 = (5x)^3 + (144^(1/3))^3#

#=(5x + 144^(1/3))(25x^2 - 5(144^(1/3))x + 144^(2/3))#

With a sum of difference of cubes we always get a single real root, meaning we cannot factor any further. So our final factorization is

#125x^3 + 144 =(5x + 144^(1/3))(25x^2 - 5(144^(1/3))x + 144^(2/3))#