Expand and simplify the square of (1 + x + x2 + x3), then subtract x cubed.

Step-by-step solution

1. 1

   Get the square by counting index pairs instead of multiplying 16 terms. In $(1+x+x^2+x^3)^2$ the coefficient of $x^k$ counts the ordered pairs $(i,j)$ with $i+j=k$ and $0\le i,j\le 3$. Counting: $k=0$ has $1$ pair, $k=1$ has $2$, $k=2$ has $3$, $k=3$ has $4$ (namely $0{+}3,1{+}2,2{+}1,3{+}0$), $k=4$ has $3$, $k=5$ has $2$, $k=6$ has $1$.

2. 2

   Write the square.

   $(1+x+x^2+x^3)^2=1+2x+3x^2+4x^3+3x^4+2x^5+x^6$

   The coefficients read the same forwards and backwards, which is expected: squaring a palindromic polynomial always gives a palindromic one.

3. 3

   Sanity-check the square at $x=1$. The base becomes $1+1+1+1=4$, so the square must be $16$; and $1+2+3+4+3+2+1=16$. Any miscount in step 1 would show up here. (For the record, $1+x+x^2+x^3=\frac{x^4-1}{x-1}$ when $x\neq 1$, so the square also equals $\frac{(x^4-1)^2}{(x-1)^2}$.)

4. 4

   Subtract $x^3$. Only one coefficient is affected — the cubic one drops from $4$ to $3$:

   $1+2x+3x^2+(4-1)x^3+3x^4+2x^5+x^6$

5. 5

   Write the result and verify at $x=2$. In descending order,

   $x^6+2x^5+3x^4+3x^3+3x^2+2x+1$

   At $x=2$ the original is $(1+2+4+8)^2-8=15^2-8=217$, and the expansion gives $64+64+48+24+12+4+1=217$. The two agree, so the expansion is correct.