Two ways to factorise a difference of sixth powers

The same polynomial can be read as a difference of squares or as a difference of cubes. The starting choice changes the number of irreducible factors obtained.

Example — Two ways to factorise $a^6-b^6$

The polynomial $a^6-b^6$ can be seen in two different ways.

Method 1 — Difference of squares first. $a^6-b^6 = (a^3)^2 - (b^3)^2 = (a^3+b^3)(a^3-b^3)$ Each factor is factorised further (sum and difference of cubes):

a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ a^3-b^3 &= (a-b)(a^2+ab+b^2) \end{aligned}$$ Result: $a^6-b^6 = \boxed{(a+b)(a^2-ab+b^2)(a-b)(a^2+ab+b^2)}$ (4 factors). **Method 2 — Difference of cubes first.** $$a^6-b^6 = (a^2)^3 - (b^2)^3 = (a^2-b^2)(a^4+a^2b^2+b^4) = (a+b)(a-b)(a^4+a^2b^2+b^4)$$ (3 factors).

Warning

Method 1 is more complete: it produces 4 irreducible factors, whereas Method 2 produces only 3 (the trinomial $a^4+a^2b^2+b^4$ does not factorise easily with the standard techniques). It is always best to start from the difference of squares.

Links

Topics: Algebraic calculus
Concepts: Difference of squares · Factorisation · Sum and difference of cubes
Skills: Factorising