With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$ I know how to expand the binomial expansion but I have no idea how to do this second part.
Hence Express $$(3 - \sqrt 2)^4$$ in the form of $$P+Q\sqrt 2$$
Where P and Q are integers, and then how to state the values of P and Q.

Here is the binomial formula with two complex numbers $a$ and $b$ :
$$(a+b)^n = \sum_{k=0}^n {n \choose k}\ a^{n-k} b^k$$ 
For $n=4$ you get :
$$\begin{array}{rcl} (a+b)^4 & = & \sum_{k=0}^4 {4 \choose k}\ a^{4-k} b^k \\ & =& {4 \choose 0}\ a^4b^0+{4 \choose 1}\ a^3 b^1+{4\choose 2}\ a^2b^2+{4 \choose 3}\ a^1b^3+{4\choose 4}\ a^0 b^4 \\
&=& a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{array}$$
Now choose $a=3$ and $b=-\sqrt 2$. 
You have $a^2=9$; $a^3=27$; and $a^4=81$.
You have $b^2=\left(-\sqrt 2\right)^2=2$; $b^3=b^2b=-2\sqrt 2$ and $b^4=\left(b^2\right)^2=2^2=4$.
Using the above formula, you get : 
$$\begin{array}{rcl}\left(3-\sqrt 2\right)^2 & =& 81+4\times 27\times\left(-\sqrt 2\right)+6\times 9\times 2+4\times 3\times \left(-2\sqrt 2\right)+4 \\
& = & 81-108\sqrt 2+108-24\sqrt 2+4 \\
& = & 193-132\sqrt 2\end{array}$$
