Factorize x^6 + 6x^3 + 5.

The task is to

Factorize x^6 + 6x^3
+ 5

My first motivation was to see that x^6 is related to x^3 by a
power of 2.

Hence I take the liberty to let y=x^3, so that x^6 would
become y^2.

The equation now becomes a quadratic equation in
y:

y^2 + 6y + 5

We can now do our usual
to factorize this quadratic equation to
become:

(y+5)(y+1)

If we replace y by
x^3 again, we would then get:

(x^3
+5).(x^3+1)

The first factor (x^3+5) can remain as it
is, since 5 is not a cubic number.

However, the 2nd factor (x^3+1)
is a little elusive.  As we know, the number 1 raised to any power still yields 1.  Hence, this 1
is a cubic number.

Also, we know that there are missing terms in the
polynomial: x^2 and x terms.  We cater for that by deliberately including the adding and
subtracting of the same term.  Then we try to regroup by "taking out" what is
common.

We now rewrite the 2nd factor as
such:

(x^3+1)

= x^3  +   x^2 - x^2  
+   x - x  + 1

= x^3  + x^2     - x^2 - x     + x +
1

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

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

Combining with the first factor, we
have:

(x^3+5) (x^2 - x + 1)
(x+1)