How do you factor completely #8x^6n - 27y^9m#?

1 Answer
George C.
Aug 30, 2016

If #m, n# are Real constants, then:

#8x^6n-27y^9m#

#=(2root(3)(n)x^2-3root(3)(m)y^3)(4root(3)(n^2)x^4+6root(3)(mn)x^2y^3+9root(3)(m^2)y^6)#

Explanation:

If we assume that #n, m# are Real constants and #x, y# are variables, then it is possible to factor this expression as a difference of cubes.

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

So we find:

#8x^6n-27y^9m#

#=(2root(3)(n)x^2)^3-(3root(3)(m)y^3)^3#

#=(2root(3)(n)x^2-3root(3)(m)y^3)((2root(3)(n)x^2)^2+(2root(3)(n)x^2)(3root(3)(m)y^3)+(3root(3)(m)y^3)^2)#

#=(2root(3)(n)x^2-3root(3)(m)y^3)(4root(3)(n^2)x^4+6root(3)(mn)x^2y^3+9root(3)(m^2)y^6)#

Related questions
See all questions in Factoring Completely
Impact of this question
1056 views around the world
You can reuse this answer
Creative Commons License