This cannot be possible ...

Consider the expression in B.

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

Thus B will always be smaller than A. We don't even need the restriction that $x$ is a positive integer. Indeed, $x$ need not even be real!

---

This is a special case of the factorization of difference of $n$ th powers: $a^n-b^n = (a-b)\left(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}\right)$ where $a=x$ and $b=1$ .

It is extremely helpful to think of this problem in base $x$ . In base $x$ , we have $x$ numbers: $0$ through $x-1$ . Each power of $x$ is assigned one place. For example, if we had $(2(x^{3}) + (4(x^{2})) + (2(x^{1})) + (3(x^{0}))$ , in base $x$ it would look like $2423_{x}$ (assuming that $x>4$ ). Note also that ${x}_{x}={10}_{x}, {x0}_{x}={100}_{x}, \cdots$ . In other words, the $x$ becomes a $0$ and gives a $1$ to the next higher place.

A. $x^{n} = {1 \overbrace{000 \cdots 000}^{\color{#3D99F6}\text{n 0's}}}_{x}$

Rearranging B, we have the following:

$(0)(x^{n})+(x-1)(x^{n-1}) + (x-1)(x^{n-2}) + (x-1)(x^{n-3}) + \cdots + (x-1)(x^{2}) + (x-1)(x^{1}) + (x-1)(x^{0})$ .

B. ${0 \overbrace{(x-1)(x-1)(x-1) \cdots (x-1)(x-1)(x-1)}^{\color{#3D99F6}\text{n (x-1)'s}}}_{x}$

If we add $1$ to B, we get:

$\begin{aligned} B={0 \overbrace{(x-1)(x-1)(x-1) \cdots (x-1)(x-1)(x)}^{\color{#3D99F6}\text{n of these terms}}}_{x} \\ &={0 \overbrace{(x-1)(x-1)(x-1) \cdots (x-1)(x)(0)}^{\color{#3D99F6}\text{n of these terms}}}_{x} \\ &={0 \overbrace{(x-1)(x-1)(x-1) \cdots (x)(0)(0)}^{\color{#3D99F6}\text{n of these terms}}}_{x} \\ \cdots \\ &={0 \overbrace{(x)(0)(0) \cdots (0)(0)(0)}^{\color{#3D99F6}\text{n of these terms}}}_{x} \\ &={1 \overbrace{000 \cdots 000}^{\color{#3D99F6}\text{n 0's}}}_{x} \\ &={x^{n}} \end{aligned}$

(Note that each time, the $x$ became a $0$ and gave a $1$ to the next higher place)

Therefore, $A=B+1 => A>B$