give an example of algebraic numbers $\alpha, \beta$ such that.... Question is to find algebraic numbers $\alpha, \beta$ such that :
$$[\mathbb{Q}(\alpha):\mathbb{Q}]>[\mathbb{Q}(\beta):\mathbb{Q}]>[\mathbb{Q}(\alpha\beta):\mathbb{Q}]$$
It is not so difficult to find algebraic numbers $\alpha, \beta$ such that :
$$[\mathbb{Q}(\alpha):\mathbb{Q}]>[\mathbb{Q}(\beta):\mathbb{Q}]$$
but then the last relation $$[\mathbb{Q}(\beta):\mathbb{Q}]>[\mathbb{Q}(\alpha\beta):\mathbb{Q}]$$is getting disturbed all the time....
Please provide some hint for me to clear this.
Thank you :)

One can argument this way. Suppose that we can find some $\alpha$ with the following properties:
1) $[\mathbb Q(\alpha)\colon \mathbb Q]=n$ and there is some prime $p$ such that $p(p+1)\mid n$.
2) The minimal polynomial $m(x)$ of $\alpha$ is of the form $f(x^{p(p+1)})$ for some $f(x)\in \mathbb Z[x]$.
Then $[\mathbb Q(\alpha^p)\colon \mathbb Q]=n/p$. In fact clearly $[\mathbb Q(\alpha^p)\colon\mathbb Q]$ is at least $n/p$, but also $\alpha^p$ is a root of $f(x^{p+1})$ which has degree $n/p$, and so is the minimal polynomial of $\alpha^p$. In the same way, one checks that $[\mathbb Q(\alpha^{p+1})\colon \mathbb Q]=n/(p+1)$ and since $\alpha^{p+1}=\alpha\cdot\alpha^p$ your example is found since $n>n/p>n/(p+1)$.
This allows you to construct a lot of examples. Pick your favourite $n$ divided by $p(p+1)$ for some prime $p$, then take any irreducible polynomial $m(x)\in \mathbb Z[x]$ of degree $n$ of the form $f(x^{p(p+1)})$ for some $f(x)\in \mathbb Z[x]$. Then any root $\alpha$ of $m$ does the job. To construct $m(x)$ one can simply use Eisenstein's criterion. 
To give an "explicit" example say that you choose $n=24=(3\cdot 4)\cdot 2$, so that $p=3$. Now fix your favourite prime number $q$ and set $m(x)=x^{24}+qx^{12}+q$. This polynomial is irreducible by Eisenstein's criterion and is of the form $f(x^{12})$ where $f(x)=x^2+qx+q$. If $\alpha$ is a root of $m$, then $[\mathbb Q(\alpha)\colon \mathbb Q]=24$, $[\mathbb Q(\alpha^3)\colon \mathbb Q]=8$ because $x^8+qx^4+q$ is the minimal polynomial of $\alpha^3$ and $[\mathbb Q(\alpha^4)\colon \mathbb Q]=6$ because the minimal polynomial of $\alpha^4$ is $x^6+qx^3+q$. Therefore $\alpha$ and $\beta=\alpha^3$ do the job.
