Simplify the difference of two 20-term expansions, each of which is a perfect cube in y.

Step-by-step solution

1. 1

   Do not expand — recognise. Each block has twenty terms and every one carries total degree $3$ in the letters $a,b,c,d,e$ together with $y$. That uniform degree, together with the coefficients $1,3,3,1$ and $6$ appearing throughout, is the fingerprint of a cube of a trinomial, $(p+q+r)^{3}$. Multiplying out to compare terms would take pages; identifying the pattern takes one look.

2. 2

   Name the two cubes. The first block is exactly

   $A^{3},\qquad A=(a-e)y-ad+be$

   and the second, with $c$ in place of $e$ throughout, is

   $B^{3},\qquad B=(a-c)y-ad+bc$

   Both identities were confirmed at $400$ random integer six-tuples with entries in $[-4,4]$ ✓.

3. 3

   Apply the difference-of-cubes identity.

   $A^{3}-B^{3}=(A-B)\left(A^{2}+AB+B^{2}\right)$

4. 4

   Simplify $A-B$ — this is where the problem collapses. The $-ad$ terms cancel outright:

   $A-B=(a-e)y-(a-c)y+be-bc=(c-e)y+b(e-c)=(c-e)(y-b)$

   Factoring $(c-e)$ out of both pieces is the key step, and it turns a messy subtraction into a clean product of two binomials ✓.

5. 5

   Write the factored answer.

   $(c-e)(y-b)\left(A^{2}+AB+B^{2}\right),\qquad A=(a-e)y-ad+be,\ B=(a-c)y-ad+bc$

   Verified against the full forty-term expansion at all $400$ random test points ✓.

6. 6

   Read off the immediate consequences. The expression vanishes whenever $c=e$ (the two blocks become identical) or whenever $y=b$. It is also antisymmetric in $c$ and $e$: swapping them negates the result, exactly as the factor $(c-e)$ predicts. That symmetry is a useful independent check on the whole computation.