How do you factor $x^{6} + 2 x^{3} + 1$ $x^6+2x^3+1$ ?

Question

6288 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-01T17:47:46

$x^{6} + 2 x^{3} + 1 = {(x + 1)}^{2} {(x^{2} - x + 1)}^{2}$ $x^6+2x^3+1 = (x+1)^2(x^2-x+1)^2$

$x^{6} + 2 x^{3} + 1$ $x^6+2x^3+1$

$= {(x^{3})}^{2} + 2 (x^{3}) + 1$ $=(x^3)^2+2(x^3)+1$

$= {(x^{3} + 1)}^{2}$ $=(x^3+1)^2$

$= {(x^{3} + 1^{3})}^{2}$ $=(x^3+1^3)^2$

$= {(x + 1)}^{2} {(x^{2} - x + 1)}^{2}$ $=(x+1)^2(x^2-x+1)^2$

Using the sum of cubes identity:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$

with $a = x$ $a=x$ and $b = 1$ $b=1$

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

How do you factor x^6+2x^3+1 x6+2x3+1x^6+2x^3+1 ?