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

May 16, 2015

${x}^{6} - 2 {x}^{3} + 1 = {\left({x}^{3}\right)}^{2} - 2 \left({x}^{3}\right) + 1$ is of the form ${y}^{2} - 2 y + 1$ where $y = {x}^{3}$.

This quadratic formula in $y$ factors as follows:

${y}^{2} - 2 y + 1 = \left(y - 1\right) \left(y - 1\right) = {\left(y - 1\right)}^{2}$

So ${x}^{6} - 2 {x}^{3} + 1 = {\left({x}^{3} - 1\right)}^{2}$

${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$

So ${x}^{6} - 2 {x}^{3} + 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right) \left(x - 1\right) \left({x}^{2} + x + 1\right)$

$= {\left(x - 1\right)}^{2} {\left({x}^{2} + x + 1\right)}^{2}$.

${x}^{2} + x + 1$ has no linear factors with real coefficients. To check this notice that it is of the form $a {x}^{2} + b x + c$, which has discriminant:

$\Delta = {b}^{2} - 4 a c = {1}^{2} - 4 \cdot 1 \cdot 1 = 1 - 4 = - 3$

Being negative, the equation ${x}^{2} + x + 1 = 0$ has no real roots.

One way of checking the answer is to substitute a value for $x$ that is not a root into both sides and see if we get the same result:

Try $x = 2$:

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

$= 64 - \left(2 \times 8\right) + 1 = 64 - 16 + 1 = 49$

Compare:

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

${1}^{2} \cdot {7}^{2} = 49$

Well that worked!

May 17, 2015

${x}^{6} - 2 {x}^{3} + 1$ is fairly easy to factor, because it is a perfect square. How do I know this? It's a trinomial in the form ${a}^{2} + 2 a b + {b}^{2}$, and all trinomials in that form are perfect squares.

This trinomial is the perfect square of $\left({x}^{3} - 1\right)$. To check my work, I'll work backwards:

$\left({x}^{3} - 1\right) \left({x}^{3} - 1\right)$

$= {x}^{6} - {x}^{3} - {x}^{3} + 1$

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

So, this trinomial has factors of $1$, ${x}^{3} - 1$, and ${x}^{6} - 2 {x}^{3} + 1$.

However, as it has been pointed out to me, $\left({x}^{3} - 1\right)$ also has factors. Since it is a binomial of the form ${a}^{3} - {b}^{3}$, it can also be written as $\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$.

So, $\left({x}^{3} - 1\right)$ factors into $\left(x - 1\right)$ and $\left({x}^{2} + x + 1\right)$, which both are prime.

The factors of ${x}^{6} - 2 {x}^{3} + 1$ are:
$1$
$x - 1$
${x}^{2} + x + 1$
${x}^{3} - 1$
${x}^{6} - 2 {x}^{3} + 1$

More specifically, the PRIME factorization of ${x}^{6} - 2 {x}^{3} + 1$ is:
${\left(x - 1\right)}^{2} {\left({x}^{2} + x + 1\right)}^{2}$