Which expression is the completely factored form of x^6−64y^3 ?

Question

2799 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-03T16:42:36

#x^6-64y^3 =(x^2-4y)(x^2+2xsqrt(y)+4y)(x^2-2xsqrt(y)+4y)#

#x^6-64y^3#

#=(x^2)^3-(4y)^3#

#=(x^2-4y))(x^4+4x^2y+16y^2)#

#=(x^2-4y)[(x^2)^2+(4y)^2+4x^2y)]#

#=(x^2-4y)[(x^2+4y)^2-2*x^2*4y+4x^2y)]#

#=(x^2-4y)[(x^2+4y)^2-8x^2y+4x^2y)]#

#=(x^2-4y)[(x^2+4y)^2-(2xsqrt(y))^2]#

#=(x^2-4y)(x^2+2xsqrt(y)+4y)(x^2-2xsqrt(y)+4y)#

Answer 2 · 2017-10-03T16:59:12

#(x^2 - 4y)(x^4+ x^(2)4y + 16y^2)#

#x^6-64y^3#

If we write this as:

#(x^2)^3-(4y)^3#: This is the difference of two cubes:

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

So:

#((x^2)^3-(4y)^3)=(x^2 - 4y)(x^4+ x^(2)4y + 16y^2)#