Proposition 5.15 from Atiyah and Macdonald: Integral Closure and Minimal Polynomial
Solution 1:
Since $x\in B$ is integral over $\mathfrak{a}$ it's algebraic over $K$. Since $x_1,\dots, x_n$ are the conjugates of $x$ (including $x$, I am guessing), then the minimal polynomial of $x$ $$f(t):=t^n+a_1t^{n-1}+\cdots+a_n$$ factors as $$f(t)=(t-x_1)(t-x_2)\cdots(t-x_n).$$ Expanding this and equating coefficients we see that \begin{align*}a_n&=(-1)^nx_1x_2\ldots x_n \\ & \vdots \\ a_2&=\sum_{i<j}x_ix_j \\ a_1&=-(x_1+x_2+\ldots+x_n).\end{align*} In particular, the $a_i$ are symmetric polynomials in the $x_i$.