How do you factor #x^6y^3 + y^9#?

1 Answer
Feb 25, 2017

Answer:

#x^6y^3+y^9 = y^3(x^2+y^2)(x^2-sqrt(3)xy+y^2)(x^2+sqrt(3)xy+y^2)#

Explanation:

The sum of cubes identity can be written:

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

So we find:

#x^6y^3+y^9 = y^3(x^6+y^6)#

#color(white)(x^6y^3+y^9) = y^3((x^2)^3+(y^2)^3)#

#color(white)(x^6y^3+y^9) = y^3(x^2+y^2)(x^4-x^2y^2+y^4)#

Then note that:

#(x^2-axy+y^2)(x^2+axy+y^2) = x^4+(2-a^2)x^2y^2+y^4#

So putting #a=sqrt(3)# we find:

#x^4-x^2y^2+y^4 = (x^2-sqrt(3)xy+y^2)(x^2+sqrt(3)xy+y^2)#

Putting it all together:

#x^6y^3+y^9 = y^3(x^2+y^2)(x^2-sqrt(3)xy+y^2)(x^2+sqrt(3)xy+y^2)#