<s>[INST] How to show that $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ is algebraic? How to show that $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$  is algebraic?
I know how to prove x = $\sqrt[3]{3} + \sqrt{2}$ and  x = $\sqrt{3} + \sqrt{2}$ , but the root doesn't disappear from $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ .
I want to make an algebraic equation that has this as an solution.
 [/INST] Here is a Simpler Approach , which intuitively shows that we will have Integer Co-efficient Equation.
Let $(x-\sqrt{2}+\sqrt{3})=(\sqrt[3]{3})$
Cubing both sides , we get :
Let $(x-\sqrt{2}+\sqrt{3})^3=(3)$
Here I will use $@$ to indicate some arbitrary Co-efficient having only Integers or $x$ or Powers. Every $@$ is a new instance , not connected to every other $@$.
In case I have to refer to the actual value , I will use $A,B,C,D$ to indicate some arbitrary Co-efficient having only Integers or $x$ or Powers.
When we expand the Cube, we will get :
$(@+@\sqrt{2}+@\sqrt{3}+@\sqrt{6})=(3)$
It looks like we got rid of $\sqrt[3]{3}$ , but introduced $\sqrt{6}$ !
Move the terms around , to get this :
$(A+B\sqrt{6})=(@\sqrt{2}+@\sqrt{3})$
Multiply both sides by $(A-B\sqrt{6})$
We get :
$(A^2-6B^2)=(@\sqrt{2}+@\sqrt{3})$ & we have got rid of the $\sqrt{6}$ !
Square both sides , to get :
$(@)=(@+@\sqrt{6})$ & we have again introduced $\sqrt{6}$ while getting rid of $\sqrt{2}$ & $\sqrt{3}$ !
Move the terms around to get this :
$(C)=(D\sqrt{6})$
Squaring both sides again , we get the Integer Co-efficient Equation which has the given root & other roots :
$(C^2)=(6D^2)$
The Degree is $3 \times 2 \times 2 \times 2 = 24$
Hence the given number is Algebraic.
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