Is $a+5^{1/3}b+5^{2/3}c$ a root of any cubic polynomial in $\mathbb{Q}$?

Question

For arbitrary $a, b, c \in \mathbb{Q}$, let $w := a + 5^{1/3} b + 5^{2/3} c$, is $w$ a root of any cubic polynomial in $\mathbb{Q}$?

I guess the cubic polynomial always exists. But I am confused about how to construct the polynomial, or how to prove the existence of the polynomial.

Answer 1

The identity used for solving cubic equations is $$(u+v)^3=u^3+v^3+3uv(u+v).$$ With $u=5^{1/3}b$ and $v=5^{2/3}c$ and $x=a+u+v$, this becomes $$(x-a)^3=5b^3+25c^3+15bc(x-a).$$ So the cubic equation you are looking for would be $x^3-3a\,x^2+(3a^2-15bc)x-a^3-5b^3-25c^3+15abc=0$.

Answer 2

It always exists because $\sqrt[3]{5}$ is algebraic over the rationals. We know that $[\mathbb{Q}(\sqrt[3]{5}):\mathbb{Q}]=3$, and $\{1,\sqrt[3]{5},\sqrt[3]{5^2}\}$ is a basis for the extension. So every $a,b,c\in\mathbb{Q}$, $w=a+b\sqrt[3]{5}+c\sqrt[3]{5^2}\in\mathbb{Q}(\sqrt[3]{5})$. Hence $[\mathbb{Q}(w):\mathbb{Q}]=3$ so exists an polynomial $f$(the minimal polynomial) whose degree is 3, such as $f(w)=0$. To construct it note that $\{1,w,w^2\}$ is a basis for the extension, now explore the linear dependence of $w^3$ with $1,w$ and $w^2$

Answer 3

More generally, let $\,\alpha\,$ be an irrational root of a rational cubic $\,\alpha^3 - \lambda \alpha^2 - \mu \alpha - \nu = 0\,$, and let $\,x = a + b \alpha + c \alpha^2\,$ for some rational $\,a, b, c\,$. Then, multiplying the first equality below twice by $\,\alpha\,$ and $\,\alpha^2\,$, and substituting $\,\alpha^3 = \lambda \alpha^2 + \mu \alpha + \nu\,$ at each step:

$$ \begin{cases} \begin{matrix} a - x &+\, b \alpha &+\, c \alpha^2 &= 0 \\ c\nu &+\, (a+c\mu-x)\alpha &+\, (b+c\lambda)\alpha^2 &= 0 \\ b\nu &+\, (c\nu + b\mu) \alpha &+\, (a + b\lambda-x) \alpha^2 &= 0 \end{matrix} \end{cases} $$

Considering the three equations as a linear system in $\,1, \alpha,\alpha^2\,$ which obviously has non-trivial solutions, the condition is for its determinant to be zero, which gives the cubic equation in $\,x\,$:

$$ \begin{vmatrix} a - x & b & c \\ c\nu & a+c\mu-x & b+c\lambda \\ b\nu & c\nu+b\mu & a+b\lambda-x \end{vmatrix} = 0 $$

OP's problem is the case $\,\alpha^3 = 5\,$, so $\,\lambda = \mu = 0, \nu = 5\,$, and the polynomial is:

$$ 0 = \begin{vmatrix} a - x & b & c \\ 5c & a-x & b \\ 5b & 5c & a-x \end{vmatrix} = - \big(x^3 - 3 a x^2 + (3 a^2 - 15 b c) x - a^3 - 5 b^3 - 25 c^3 + 15 a b c\big) $$

(Note: the above is equivalent to deriving by hand the resultant $\small{\text{res}_{\alpha}(\alpha^3 - 5, x - a - b \alpha - c \alpha^2)}$.)