How do you factor #x^6+2x^3+1 #?

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Answer 1 · 2015-07-01T17:47:46

#x^6+2x^3+1 = (x+1)^2(x^2-x+1)^2#

#x^6+2x^3+1#

#=(x^3)^2+2(x^3)+1#

#=(x^3+1)^2#

#=(x^3+1^3)^2#

#=(x+1)^2(x^2-x+1)^2#

Using the sum of cubes identity:

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

with #a=x# and #b=1#

#(x^2-x+1)# has no simpler factors with real coefficients.