Expressing a number field in a simpler way 
Express $\mathbb{Q}(\sqrt{3},\sqrt[3]{5})$ in the form $\mathbb{Q}(\theta)$.
Hint: Let $f,g$ be the minimal polynomials of $\sqrt{3}$ and $\sqrt[3]{5}$, respectively. Factor $f$ and $g$ in $\mathbb{C}$ to obtain expressions of the form $$f(x)=\prod\limits_{j=1}^n (x-\alpha_j),g(x)=\prod\limits_{k=1}^m (x-\beta_k)$$
where $\alpha_1=\sqrt{3}$ and $\beta_1=\sqrt[3]{5}$. Choose $c \in \mathbb{Q}$\ $\lbrace 0 \rbrace$ so that $$\alpha+c\beta \notin \lbrace \alpha_j+c\beta_k|1\leq j \leq n, 2 \leq k \leq m \rbrace$$ It can then be shown that $\mathbb{Q}(\sqrt{3},\sqrt[3]{5})=\mathbb{Q}(\sqrt{3}+c\sqrt[3]{5})$

Following the hint, I think $c=1$ works. Now I try to prove that $\mathbb{Q}(\sqrt{3},\sqrt[3]{5}) \subseteq \mathbb{Q}(\sqrt{3}+\sqrt[3]{5})$ (the other containment is easy), but I'm having trouble with this last step.
 A: If you were working on an exercise, I’m not sure that what I’m about to show you will satisfy the requirements, though it certainly does show that $\Bbb Q(\sqrt3+\sqrt[3]5)$ is the compositum of the two given fields.
Look first at the field $\Bbb Q(\sqrt3)$. Its ring of integers is $\Bbb Z[\sqrt3]$, and this is a unique factorization domain in which $5$ is still a prime. Now consider the polynomial $F(X)=X^3-5$ over your quadratic field. Since $5$ is prime, Eisenstein applies, and this cubic is irreducible. This verifies that $[\Bbb Q(\sqrt2,\sqrt[3]5)\colon\Bbb Q]=6$, as you already knew.
Now form $G(X)=F(X-\sqrt3)$, a polynomial that has your quantity $\sqrt3+\sqrt[3]5$ as a root, and which is still irreducible over $\Bbb Q(\sqrt3)$.
Next, form $\overline G$, the conjugate polynomial, it’s just what you get by replacing 
$\sqrt3$ by $-\sqrt3$ wherever it appears in $G$. The polynomial $G\overline G$ clearly has $\Bbb Q$-coefficients (in fact, it’s $X^6 -9X^4-10X^3 +27X^2-90X-2$), call it $H(X)$, and $H$ has $\sqrt3+\sqrt[3]5$ as a root.
The only question is whether $H$ is $\Bbb Q$-irreducible. It would be reducible if the field generated by $\sqrt3+\sqrt[3]5$ was smaller than your sextic field. But now I’m going to argue that $H$ is $\Bbb Q$-irreducible. If it weren’t, its factorization would also be a factorization into $Q(\sqrt3)$-polynomials. But we know how $H$ factors into irreducibles over the quadratic field, it’s $H=G\overline G$, and this is the only factorization of $H$ over the quadratic field, because $G$ and its conjugate are irreducible there. So there is no such factorization over $\Bbb Q$, and your irrational quantity is indeed of degree six over the rationals.
A: Denote
$$
a=\sqrt{3},\quad b=\sqrt[3]{5}, \quad x = a+b.
$$
Let's express $1,x,x^2,x^3,x^4,x^5$ as linear combinations of $1,a,b,ab,b^2,ab^2$:
$$
\begin{array}{lll}
1= 1; & \\
x=  a+b; & \\
x^2= (a+b)^2 & = a^2+2ab+b^2 & = 3+2ab+b^2;\\
x^3= (a+b)^3 & = a^3+3a^2b+3ab^2+b^3 & = 3a+9b+3ab^2+5;\\
x^4= (a+b)^4 & = a^4+4a^3b+6a^2b^2+4ab^3+b^4 & = 9+12ab+18b^2+20a+5b;\\
x^5= (a+b)^5 & = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 & = 9a+45b+30ab^2+150+25ab+5b^2.
\end{array}
$$
Following this way, one can express:
$$
\left(
\begin{array}{c}
1 \\ x \\ x^2 \\ x^3 \\ x^4 \\ x^5 
\end{array}
\right)
=
\left(
\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 \\
3 & 0 & 0 & 2 & 1 & 0 \\
5 & 3 & 9 & 0 & 0 & 3 \\
9 & 20 & 5 & 12 & 18 & 0 \\
150 & 9 & 45 & 25 & 5 & 30
\end{array}
\right)
\cdot
\left(
\begin{array}{c}
1 \\ a \\ b \\ ab \\ b^2 \\ ab^2 
\end{array}
\right),
$$
or, say,
$$
X = M\cdot A.
$$
From this,
$$
A = M^{-1}\cdot X;
$$
in particular,
$$
\sqrt{3} = a = \frac{1}{2403}\left(-4095+3015x-1170x^2-720x^3+45x^4+72x^5\right),
$$
$$
\sqrt[3]{5} = b = \frac{1}{2403}\left(4095-612x+1170x^2+720x^3-45x^4-72x^5\right),
$$
(where $x=\sqrt{3}+\sqrt[3]{5}$).
A: If you look at the proof of the primitive element theorem--at least the constructive version--you see that the finitely many exceptions are spelled out exactly the $c$ so that
$$\theta=\sqrt{3}+c\sqrt[3]{5}=\alpha_i+c\beta_j,\;\text{ for some } 1\le i\le n, 2\le j\le m$$
But you know the other roots of $x^2-3$ and $x^3-5$ explicitly as elements of $\Bbb C$, and you can tell that if $2\le j\le k$ then the number $\alpha_i+c\beta_j$ is a real number--$\alpha_i$ is always real-- plus a purely complex (i.e. not real) number--$\zeta_3\sqrt[3]{5}=\beta_j$ when $2\le j\le 3$ for some $\zeta_3$ a primitive third root of $1$. We conclude no $\alpha_i+\beta_j$ (here I have not written $c$ because we have chosen $c=1$) is real, hence no $\alpha_i+\beta_j=\sqrt{3}+\sqrt[3]{5}$ for $2\le j\le 3$.
