<s>[INST] Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $ 
Find the polynomial $P$ of smallest degree with rational coefficients
  and leading coefficient $1$ such that $$ P(49^{1/3}+7^{1/3})=4 $$  
(Source:NYSML)

My attempt
Let $$ 49^{1/3}+7^{1/3}=x$$
Then,
\begin{array}
((49^{1/3})^3 &=(x-7^{1/3})^{3}  \\
49 &=x^3-7-3x^2\cdot 7^{1/3}+3x \cdot 7^{2/3} \\
\end{array}
So I have $$x^3+x^2 (-3\cdot 7^{1/3})+x(3\cdot 7^{2/3})-56 =0 $$
Define $$Q(x)=x^3+x^2 (-3\cdot 7^{1/3})+x(3\cdot 7^{2/3})-56 $$
making the substitution $x=7^{1/3} \cdot t $,
I have $$F(t)=\cfrac{Q(7^{1/3} \cdot t)}{7}=t^3+t^2(-21)+21t-56$$
Now I am quite stuck on how I should continue...
 [/INST] Your first attempt (cubing $49^{1/3}+7^{1/3}$) came close to finishing. Let $a$ be our number. Cubing, we find that
$$a^3=56+3(7^{5/3}+7^{4/3})=56+21a.$$
So if $Q(x)$ is the polynomial $x^3-21x-56$, then $Q(a)=0$. Dealing with the fact that we want $P(a)=4$ is now easy.
We sketch a proof that this polynomial $P(x)$ has minimal degree. First show that $a$ is not a root of a polynomial of degree $1$ with rational coefficients. This can be handled by the Rational Root Theorem. To show that $a$ is not a root of a polynomial of degree $2$ with rational coefficients, suppose that $W(x)$ is such a polynomial. Then the remainder $R(x)$ when we divide $P(x)$ by $W(x)$ has degree $\le 1$, and $R(a)=0$, which is impossible.
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