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:
- \(a^2-b^2 = (a+b)(a-b)\),
- \(a^3-b^3 = (a-b)(a^2+ab+b^2)\),
- \(a^3+b^3 = (a+b)(a^2-ab+b^2)\),
- \((a+b)^3 = a^3+3a^2b+3ab^2+b^3\),
- \((a-b)^3 = a^3-3a^2b+3ab^2-b^3\).
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 \[\begin{align*} (x+y)^6 &= x^6 + \binom{6}{1} x^5y + \dotsb + \binom{6}{5} xy^5 + y^6\\ &= x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\\ (x-y)^6 &= x^6 - 6x^5y + 15x^4y^2 - 20x^3y^3 + 15x^2y^4 - 6xy^5 + y^6, \end{align*}\] so that \[\begin{align*} (x+y)^6 - (x-y)^6 &= 12x^5y + 40x^3y^3 + 12xy^5 \\ &= 4xy(3x^4+10x^2y^2+3y^4)\\ &= 4xy(x^2+3y^2)(3x^2+y^2). \end{align*}\]
