Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:   


  Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ . 
  
  Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
  
  $(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$ 
  
  Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$ 
  
  Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$. 
  
  Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
  So degree of $p(x)$ is atleast $4$.
  
  But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .
  
  So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$


But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$ 

Kindly help with this.
That fact comes from Galois Theory. Under the assumption that $K/F$ is Galois, the (monic) minimal polynomial $f$ of an element $\alpha \in K$ (over $F$) is of the form
$$f(x) = \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma \alpha).$$
This implies that $\sigma \alpha$ are all roots of the minimal polynomial for $\alpha$.

In your case, let $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ and $F = \mathbb{Q}$. Then $K/F$ is Galois and there are 4 automorphisms in $\text{Gal}(K/F)$: namely


identity on $K$
the one that sends $\sqrt{2} \mapsto -\sqrt{2}$ keeping $\sqrt{3}$ fixed: this automorphism sends $\sqrt{2} + \sqrt{3}$ to $-\sqrt{2} + \sqrt{3}$; hence $-\sqrt{2} + \sqrt{3}$ must be a root of $p(x)$.
the one that keeps $\sqrt{2}$ fixed and sends $\sqrt{3} \mapsto -\sqrt{3}$: like above, we deduce $\sqrt{2} - \sqrt{3}$ is root of minimal polynomial $p(x)$
the one that negate both roots: finally $- \sqrt{2} - \sqrt{3}$ must also be a root.
Assume that $m$ and $n$ are square-free coprime integers. Then, if we prove that $Q(\sqrt m +\sqrt n) = Q(\sqrt m, \sqrt n)$, we can conclude that that $f(x) = x^4 -2(m+n)x +(m-n)^2$ must be the minimal polynomial of $\sqrt m +\sqrt n$ since $Q(\sqrt m +\sqrt n)$ is a four-dimensional $Q$-vector space. But if $\xi = (\sqrt m + \sqrt n)$, then ${\frac {{\xi}^2-(m+n)} {2}}={\sqrt m}{\sqrt n}$. Thus, $({\frac {{\xi}^2-(m+n)} {2}})\xi = m{\sqrt n}+n{\sqrt m}$. Hence, $$({\frac {{\xi}^2-(m+n)} {2}})\xi -n{\xi} = (m-n){\sqrt n},$$ and $$({\frac {{\xi}^2-(m+n)} {2}})\xi -m{\xi} = (n-m){\sqrt n}.$$ Thus, $Q(\sqrt m +\sqrt n)$ contains $\{\sqrt m, \sqrt n \}$, and so $Q(\sqrt m +\sqrt n)\supset Q(\sqrt m ,\sqrt n)$. The reverse set inclusion is immediate. The case $m =2$, and $n = 3$ is of course what inspired this train of thought which is hopefully correct.