Find the following determinant $$\left| \begin{matrix}x_1&a&a&...&a\\ a&x_2&a&...&a\\a&a&x_3&...&a\\\vdots&\vdots&\vdots&\vdots&\vdots\\a&a&a&...&x_n \end{matrix} \right|$$
I decided to subtract the last row from all other rows to get $$\left| \begin{matrix}x_1-a&0&0&...&a-x_n\\ 0&x_2-a&0&...&a-x_n\\0&0&x_3-a&...&a-x_n\\\vdots&\vdots&\vdots&\vdots&\vdots\\a&a&a&...&x_n \end{matrix} \right|$$
How do I continue from here?
Let $g : x \mapsto \displaystyle \prod_{k=1}^n (x_k-x)$. For any $b,x \in \mathbb{R}$, let $$A_b(x)=\begin{pmatrix}x_1+x&a+x&a+x&...&a+x\\ b+x&x_2+x&a+x&...&a+x\\b+x&b+x&x_3+x&...&a+x\\\vdots&\vdots&\vdots&\vdots&\vdots\\b+x&b+x&b+x&...&x_n+x \end{pmatrix}$$
It is easy to see that the function $f : x \mapsto \det(A_b(x))$ is a polynomial function of degree $1$ (you can see this by substracting the first column to all the other columns, and then develop the determinant w.r.t the first column).
But one has directly $\displaystyle f(-a)=g(a)$ and $\displaystyle f(-b)=g(b)$, so for every $x \in \mathbb{R}$, and $b \neq a$, one has $$f(x)=\dfrac{f(-b)-f(-a)}{-b+a}(x+a)+f(-a) = -\dfrac{g(b) - g(a)}{b-a}(x+a) +g(a)$$
which gives (still for $b \neq a$) $$\det(A_b(0))=f(0) =g(a) -a\dfrac{g(b) - g(a)}{b-a} $$
Now, since $b \mapsto \det(A_b(0))$ is polynomial, then it is continuous, so $\displaystyle \det(A_a(0))=\lim_{b \rightarrow a} \det(A_b(0))$.
One gets $$\det(A_a(0))=g(a)-g'(a)a$$
i.e. finally $$\boxed{\begin{vmatrix}x_1&a&a&...&a\\ a&x_2&a&...&a\\a&a&x_3&...&a\\\vdots&\vdots&\vdots&\vdots&\vdots\\a&a&a&...&x_n \end{vmatrix} = \prod_{k=1}^n (x_k-a) + a \sum_{k=1}^n \prod_{l=1, l\neq k}^n (x_l-a)}$$
I just wanted to present an alternative approach using the matrix determinant lemma. Your matrix can be written in the form $D + axx^T$, where $D$ is the diagonal matrix and $x$ the column-vector given by $$ D = \pmatrix{x_1 - a\\ & \ddots \\ && x_n - a}, \quad x = \pmatrix{1 \\ \vdots \\ 1}. $$ We have $$ \det(D + axx^T) = (1 + ax^TD^{-1}x)\det(D). $$ We calculate $$ \det(D) = \prod_{i = 1}^n (x_i - a),\quad x^TD^{-1}x = \sum_{i=1}^n \frac 1{x_i - a}. $$ Thus, we have the following formula for the determinant: $$ \det(D + axx^T) = \left(1 + a\sum_{i=1}^n \frac 1{x_i - a} \right)\prod_{i = 1}^n (x_i - a), $$ which can be further simplified to yield TheSilverDoe's answer.
Let $D_a(x_1,\dots, x_n)$ be the determinant, then
$$ D_a(x_1,\dots,x_n) = (x_1-a) D_a(x_2,\dots,x_n)+(-1)^n (a-x_n)(-1)^{n-1} a \prod\limits_{i=2}^{n-1} (x_i - a), $$ as follows from Laplace expansion. This simplifies into $$ = (x_1 -a)D_a(x_2,\dots,x_n) + a \prod\limits_{i=2}^{n} (x_i-a). $$ With the base case $D_a(x_n) = x_n$. After expanding $D_a(x_2,\dots,x_n)$ up to $D_a(x_k,\dots,x_n)$ we get $$ = \prod\limits_{i=1}^{k-1} (x_i-a) D_a(x_k,\dots,x_n) +a\sum\limits_{j=1}^{k-1} \prod\limits_{i\neq j} (x_i-a), $$ which can be proven by induction. Expanding it all the way up to $D_a(x_n)$, we get $$ =x_n \prod\limits_{i=1}^{n-1} (x_i-a) + a\sum\limits_{j=1}^{n-1} \prod\limits_{i\neq j}(x_i-a). $$ This then simplifies into a symmetric form from @TheSilverDoe answer: $$ = \prod\limits_{i=1}^n (x_i-a) + a \sum\limits_{j=1}^n \prod\limits_{i\neq j} (x_i-a), $$ which, assuming $a \not\in\{x_1,\dots,x_n\}$ can be further refactored as $$ =\left(1+\sum\limits_{j=1}^n\frac{a}{x_j-a}\right) \prod\limits_{i=1}^n(x_i-a), $$
or as $P(a) - a P'(a)$ for $P(a) = \prod\limits_{i=1}^n (x_i-a)$.
