Hint $\ $ A root $\alpha$ of a polynomial $\,f(x)\,$ divides its constant term $\,\alpha\mid f(0)$. Over a field, if $\,\alpha\ne 0\,$ then by cancelling $\,\alpha^k$ we may assume that $\,f(0)\ne 0\,$ so $\,f(0)\,$ is a unit, thus so is its factor $\,\alpha.$

Remark $\ $ The above rearrangement of a minimal polynomial to show that a root divides the constant term is simply a generalization of rationalizing denominators. Indeed a quadratic $\,\alpha\,$ has minimal polynomial $\,f(x) = (x-\alpha)(x-\alpha') = x^2-tx + n,\,$ $\, t = \alpha+\alpha',\,$ $\, n = \alpha\alpha',\,$ hence rearranging $\,\alpha^2 - t\alpha + n = 0\,$ yields $\,\alpha(t-\alpha) = n,\,$ therefore $\,\alpha((t-\alpha)/n) = 1,\,$ hence $\,\alpha^{-1} = (t-\alpha)/n = \alpha'/(\alpha\alpha'),\,$ precisely the result of rationalizing the denominator of $\,1/\alpha.\,$

The same method works for higher degree algebraic numbers by multiplying the denominator by all other conjugates, i.e. using $\,\alpha\mid \alpha\alpha'\cdots \alpha^{(n)} = N(\alpha),\,$ for $\,N = $ norm. This works because the norm is a nonzero element of a field, hence is invertible. Essentially this exploits the norm homomorphism in order to reduce the problem of division in $\,K(\alpha)\,$ to the simpler problem of division in $\,K.$ The key idea of the hint is that we can find such a simpler multiple of $\,\alpha\,$ already in the constant term of the minimal polynomial.

Hint $\ $ A root $\alpha$ of a polynomial $\,f(x)\,$ divides its constant term $\,\alpha\mid f(0)$. Over a field, if $\,\alpha\ne 0\,$ then by cancelling $\,\alpha^k$ we may assume that $\,f(0)\ne 0\,$ so $\,f(0)\,$ is a unit, thus so is its factor $\,\alpha.$

Remark $\ $ The above rearrangement of a minimal polynomial to show that a root divides the constant term is simply a generalization of rationalizing denominators. Indeed a quadratic $\,\alpha\,$ has minimal polynomial $\,f(x) = (x-\alpha)(x-\alpha') = x^2-tx + n,\,$ $\, t = \alpha+\alpha',\,$ $\, n = \alpha\alpha',\,$ hence rearranging $\,\alpha^2 - t\alpha + n = 0\,$ yields $\,\alpha(t-\alpha) = n,\,$ hence $\,\alpha((t-\alpha)/n) = 1,\,$ therefore $\,\alpha^{-1} = \dfrac{t-\alpha}n = \dfrac{\alpha'}{\alpha\alpha'},\,$ precisely the result of rationalizing the denominator of $\,\dfrac{1}{\alpha}.\,$

The same method works for higher degree algebraic numbers by multiplying the denominator by all other conjugates, i.e. using $\,\alpha\mid \alpha\alpha'\cdots \alpha^{(n)} = N(\alpha),\,$ for $\,N = $ norm. This works because the norm is a nonzero element of a field, hence is invertible. Essentially this exploits the norm homomorphism in order to reduce the problem of division in $\,K(\alpha)\,$ to the simpler problem of division in $\,K.$ The key idea of the hint is that we can find such a simpler multiple of $\,\alpha\,$ already in the constant term of the minimal polynomial.

Hint $\ $ A root $\alpha$ of a polynomial $\,f(x)\,$ divides its constant term $\,\alpha\mid f(0)$. Over a field, if $\,\alpha\ne 0\,$ then by cancelling $\,\alpha^k$ we may assume that $\,f(0)\ne 0\,$ so $\,f(0)\,$ is a unit, thus so is its factor $\,\alpha.$

Remark $\ $ This can be viewed as a generalization of $\,\alpha\mid N(\alpha),\,$ for $N = $ norm, which is a generalization of rationalizing denominators.

Hint $\ $ A root $\alpha$ of a polynomial $\,f(x)\,$ divides its constant term $\,\alpha\mid f(0)$. Over a field, if $\,\alpha\ne 0\,$ then by cancelling $\,\alpha^k$ we may assume that $\,f(0)\ne 0\,$ so $\,f(0)\,$ is a unit, thus so is its factor $\,\alpha.$

Remark $\ $ The above rearrangement of a minimal polynomial to show that a root divides the constant term is simply a generalization of rationalizing denominators. Indeed a quadratic $\,\alpha\,$ has minimal polynomial $\,f(x) = (x-\alpha)(x-\alpha') = x^2-tx + n,\,$ $\, t = \alpha+\alpha',\,$ $\, n = \alpha\alpha',\,$ hence rearranging $\,\alpha^2 - t\alpha + n = 0\,$ yields $\,\alpha(t-\alpha) = n,\,$ therefore $\,\alpha((t-\alpha)/n) = 1,\,$ hence $\,\alpha^{-1} = (t-\alpha)/n = \alpha'/(\alpha\alpha'),\,$ precisely the result of rationalizing the denominator of $\,1/\alpha.\,$

The same method works for higher degree algebraic numbers by multiplying the denominator by all other conjugates, i.e. using $\,\alpha\mid \alpha\alpha'\cdots \alpha^{(n)} = N(\alpha),\,$ for $\,N = $ norm. This works because the norm is a nonzero element of a field, hence is invertible. Essentially this exploits the norm homomorphism in order to reduce the problem of division in $\,K(\alpha)\,$ to the simpler problem of division in $\,K.$ The key idea of the hint is that we can find such a simpler multiple of $\,\alpha\,$ already in the constant term of the minimal polynomial.