Factor x^105 - 1.

Step-by-step solution

1. 1

   Note that parity is irrelevant here. Unlike a sum of powers, a difference $x^{n}-1$ always has $x-1$ as a factor, for every positive integer $n$. The factor theorem shows why in one line: substituting $x=1$ gives $1^{105}-1=0$ ✓ regardless of the exponent.

2. 2

   Apply the difference-of-powers identity. With $a=x$, $b=1$ and $n=105$:

   $a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)$

   All signs in the cofactor are positive — this is the structural difference from the sum case, where they alternate.

3. 3

   Write the factorisation.

   $x^{105}-1=(x-1)\left(x^{104}+x^{103}+x^{102}+\cdots+x+1\right)$

   The cofactor is the finite geometric sum $\displaystyle\sum_{k=0}^{104}x^{k}$, which is exactly why the identity is equivalent to the geometric series formula $\displaystyle\sum_{k=0}^{n-1}x^{k}=\frac{x^{n}-1}{x-1}$ for $x\neq1$.

4. 4

   Verify by telescoping and numerically. Multiplying out, $x\cdot\sum x^{k}$ gives degrees $1$ to $105$ and $-1\cdot\sum x^{k}$ gives degrees $0$ to $104$; every degree from $1$ to $104$ cancels, leaving $x^{105}-1$ ✓. Exact integer evaluation at $x=2,3,5,-2,-3$ agrees in all cases ✓.

5. 5

   Push further using the divisors of 105. Because $105=3\times5\times7$, the expression also splits along each divisor — for example

   $x^{105}-1=\left(x^{5}\right)^{21}-1=\left(x^{5}-1\right)\left(x^{100}+x^{95}+\cdots+1\right)$

   In general $x^{d}-1$ divides $x^{105}-1$ for every divisor $d$ of $105$, namely $1,3,5,7,15,21,35,105$. The complete factorisation over $\mathbb{Q}$ is the product of the cyclotomic polynomials $\Phi_{d}(x)$ over exactly those eight divisors.