<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] A fast way to get the polynomial is to observe that
$$(x+y)^3=x^3+3x^2y+3xy^3+y^3=x^3+y^3+3xy(x+y)  (**)$$
Now with your $a=49^{1/3}+7^{1/3}$ setting $x=49^{1/3}, y=7^{1/3}$ you have 
$$x^3=49, y^3=7, xy=7$$
Replace those in $(**)$ and you are done.
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