What is the solution of this algebraic problem?

Question

Let $a, b, n, X$ and $Y \in \mathbb{N}$. Find $X$ and $Y$ such that

$(a+b)^{n} = X + Yb$.

Answer 1

Using the binomial theorem, we have:

\begin{align*} (a + b)^n &= \binom{n}{n} a^n b^0 + \binom{n}{n - 1} a^{n - 1} b + \dots + \binom{n}{1} a b^{n - 1} + \binom{n}{0} a^0 b^n \\ &= a^n + \binom{n}{n - 1} a^{n - 1} b + \dots + \binom{n}{1} a b^{n - 1} + b^n \end{align*}

Let $Y = b^{n - 1}$ and $X = a^n + \binom{n}{n - 1} a^{n - 1} b + \dots + \binom{n}{1} a b^{n - 1}$.

Then,

\begin{align*} X + Yb &= a^n + \binom{n}{n - 1} a^{n - 1} b + \dots + \binom{n}{1} a b^{n - 1} + (b^{n - 1}) b \\ &= \binom{n}{n} a^n b^0 + \binom{n}{n - 1} a^{n - 1} b + \dots + \binom{n}{1} a b^{n - 1} + \binom{n}{0} a^0 b^n \\ &= (a + b)^n \end{align*}

Therefore, $X + Yb = (a + b)^n$.