How do you factor $x^{6} - 26 x^{3} - 27$ $x^6-26x^3-27$ ?

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Answer 1 · 2015-02-05T04:18:42

First let's notice that $x^{6} = {(x^{3})}^{2}$ $x^6=(x^3)^2$

Let $r = x^{3}$ $r=x^3$

Using substitution we have

$r^{2} - 26 r - 27$ $r^2-26r-27$

We now look for the factors of $- 27$ $-27$ that will combine (add) to make $- 26$ $-26$

We find that $- 27$ $-27$ and $1$ $1$ multiply to be $- 27$ $-27$ and add to be $- 26$ $-26$ . This gives us

$(r - 27) (r + 1)$ $(r-27)(r+1)$

Now substitute $x^{3}$ $x^3$ back in for $r$ $r$

$(x^{3} - 27) (x^{3} + 1)$ $(x^3-27)(x^3+1)$

We should recognize this as a difference of cubes and a sum of cubes.

$(x - 3) (x^{2} + 3 x + 9) (x + 1) (x^{2} - x + 1)$ $(x-3)(x^2+3x+9)(x+1)(x^2-x+1)$

How do you factor x^6-26x^3-27x6−26x3−27x^6-26x^3-27?