<s>[INST] 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.
 [/INST] Using the binomial theorem
$$
\begin{align*}
(3-\sqrt{2})^4 &= \sum_{k = 0}^4 \binom{4}{k} (3)^k (-\sqrt{2})^{4-k}\\
&= 3^0\binom{4}{0}(-\sqrt{2})^4 + (3^1) \binom{4}{1} (-\sqrt{2})^3 + (3^2)\binom{4}{2}(-\sqrt{2})^2 \\
&\quad + (3^3)\binom{4}{3}(-\sqrt{2}) + (3^4)\binom{4}{4}(-\sqrt{2})^0\\
&= (1)(1)(4) + (3)(4)(-2\sqrt{2}) + (9)(6)(2) + (27)(4)(-\sqrt{2}) + (81)(1)(1)\\
&= 193 - 132\sqrt{2}
\end{align*}
$$
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