How do you factor #x^6+10x^3+25#?

1 Answer
Sep 2, 2016

Answer:

#x^6+10x^3+25 = (x^3+5)^2 = (x+root(3)(5))^2(x^2-(root(3)(5))x+root(3)(25))^2#

Explanation:

The sum of cubes identity can be written:

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

Use this with #a=x# and #b=root(3)(5)# as follows:

#x^6+10x^3+25 = (x^3)^2+2(5)(x^3)+(5)^2#

#color(white)(x^6+10x^3+25) = (x^3+5)^2#

#color(white)(x^6+10x^3+25) = (x^3+(root(3)(5))^3)^2#

#color(white)(x^6+10x^3+25) = ((x+root(3)(5))(x^2-x(root(3)(5))+(root(3)(5))^2))^2#

#color(white)(x^6+10x^3+25) = (x+root(3)(5))^2(x^2-(root(3)(5))x+root(3)(25))^2#