Simplify (x + 1)(x + 1)(x + 1)(x + 1) - (x + 3)(x + 3)(x + 3)(x + 3) - 272.

Step-by-step solution

1. 1

   Rewrite the repeated products as fourth powers. The four identical factors in each group are just powers:

   $(x+1)(x+1)(x+1)(x+1)=(x+1)^4,\qquad (x+3)(x+3)(x+3)(x+3)=(x+3)^4$

2. 2

   Expand each with the binomial theorem. Using coefficients $1,4,6,4,1$:

   $(x+1)^4=x^4+4x^3+6x^2+4x+1$

   $(x+3)^4=x^4+4(3)x^3+6(9)x^2+4(27)x+81=x^4+12x^3+54x^2+108x+81$

   The powers of $3$ grow quickly — $3,9,27,81$ — and that is where arithmetic slips happen.

3. 3

   Subtract term by term, distributing the minus over the whole second expansion.

   $x^4-x^4=0,\quad 4x^3-12x^3=-8x^3,\quad 6x^2-54x^2=-48x^2,\quad 4x-108x=-104x$

   The leading $x^4$ terms always cancel when two monic fourth powers are subtracted, so the result drops to degree $3$.

4. 4

   Handle the constants, including the $-272$.

   $1-81-272=-352$

   so the simplified expression is

   $-8x^3-48x^2-104x-352$

5. 5

   Factor out the common $-8$ and verify. Each coefficient is divisible by $8$:

   $-8\left(x^3+6x^2+13x+44\right)$

   Checking at $x=0$: the original is $1-81-272=-352$ and the simplified form gives $-352\;\checkmark$. At $x=1$: the original is $16-256-272=-512$, and $-8-48-104-352=-512\;\checkmark$. The cubic factor has no rational roots among the divisors of $44$, so this is as far as integer factoring goes.