2011/067) factor $a^6 + 18a^3 + 125$

the above is not difficult though it looks so

we realize $a^6 = (a^2)^3$ and $125 = 5^3$

now if we can split (is it possible) 18 so that $18 = x^3 + y$ and  $-y = 15 x$  (which is true x = 3 and y = -45)

we get $a^6 + 18a^3 + 125 = a^6 + 27 a^3 - 45 a^3 + 125$
$= (a^2)^3 + (3a)^2 + 5^3 - 3(a^2)(3a) 5$

and using $x^3 + y^3 + z^3 = (x+y+z)(x^2+y^2+z^2- xy-yz-zx)$
we get
$(a² - 3a + 5)((a^4) + 9a^2 + 25 + 3(a^3) + 15a - 5a²)$

$= (a² - 3a + 5)((a^4) + 3(a^3) + 4a² + 15a + 25)$