Can we factorise $(x+y)^6 - (x-y)^6$ completely?

Can we factorise $(x+y)^6 - (x-y)^6$ completely?

Question

Factorise completely

$$\begin{equation*} (x+y)^6 - (x-y)^6. \end{equation*}$$

Suggestion

Can we factorise simpler things? $6=2\times3$, so how about thinking about $a^2-b^2$? Or $a^3 - b^3$? Or even $a^3 + b^3$?

Or else we might wonder if, while it’s usually a bad idea, it’s worth trying to expand…

Solution

Since $6=2\times3$, we can think of this expression as $((x+y)^2)^3-((x-y)^2)^3$ or as $((x+y)^3)^2-((x-y)^3)^2$.

There are then several identities that are helpful to us here:

1. $a^2-b^2 = (a+b)(a-b)$,

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1. $a^3-b^3 = (a-b)(a^2+ab+b^2)$,

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1. $a^3+b^3 = (a+b)(a^2-ab+b^2)$,

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1. $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$,

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1. $(a-b)^3 = a^3-3a^2b+3ab^2-b^3$.

We can check these identities by multiplying out the brackets. Identities (4) and (5) are examples of the binomial theorem.

If we start with $(x+y)^6-(x-y)^6=((x+y)^3)^2-((x-y)^3)^2$, we can use (1) to get $(x+y)^6 - (x-y)^6 = \bigl((x+y)^3-(x-y)^3\bigr)\bigl((x+y)^3+(x-y)^3\bigr).$

Now we can expand the inner brackets using the binomial theorem to get

$$\begin{align*} (x+y)^6 - (x-y)^6 &= \bigl((x^3+3x^2y+3xy^2+y^3)-(x^3-3x^2y+3xy^2-y^3)\bigr)\\ &\qquad\times \bigl((x^3+3x^2y+3xy^2+y^3)+(x^3-3x^2y+3xy^2-y^3)\bigr)\\ &= (6x^2y+2y^3)(2x^3+6xy^2)\\ &= 4xy(3x^2+y^2)(x^2+3y^2). \end{align*}$$

We cannot factorise this any further, as each bracket is the sum of two squares.

Alternatively, if we start with $(x+y)^6-(x-y)^6=((x+y)^2)^3-((x-y)^2)^3$, we can use (2) to get $(x+y)^6 - (x-y)^6 = \bigl((x+y)^2-(x-y)^2\bigr)\bigl((x+y)^4+(x+y)^2(x-y)^2+(x-y)^4\bigr).$

Expanding the inner brackets using the binomial theorem and then simplifying will eventually give the same result, but it seems like a lot more work.

As a third route, we can simply expand the brackets in

$(x+y)^6 - (x-y)^6$

directly using the

binomial theorem

. We have

$$\begin{align*} (x+y)^6 &= x^6 + \binom{6}{1} x^5y + \dots b + \binom{6}{5} xy^5 + y^6\\ &= x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\\ (x-y)^6 &= x^6 - 6x^5y + 15x^4y^2 - 20x^3y^3 + 15x^2y^4 - 6xy^5 + y^6, \end{align*}$$

so that

$$\begin{align*} (x+y)^6 - (x-y)^6 &= 12x^5y + 40x^3y^3 + 12xy^5 \\ &= 4xy(3x^4+10x^2y^2+3y^4)\\ &= 4xy(x^2+3y^2)(3x^2+y^2). \end{align*}$$