If $a=3-2 \sqrt{2}$, find the value of $a^{2}-\frac{1}{a^{2}}$
$a=3-2 \sqrt{2}$
$\Rightarrow a^{2}=(3-2 \sqrt{2})^{2}$
$\Rightarrow a^{2}=9+8-12 \sqrt{2}$
$\Rightarrow a^{2}=17-12 \sqrt{2}$ ..........(1)
$\therefore \frac{1}{a^{2}}=\frac{1}{17-12 \sqrt{2}}$
$\Rightarrow \frac{1}{a^{2}}=\frac{1}{17-12 \sqrt{2}} \times \frac{17+12 \sqrt{2}}{17+12 \sqrt{2}}$
$\Rightarrow \frac{1}{a^{2}}=\frac{17+12 \sqrt{2}}{17^{2}-(12 \sqrt{2})^{2}}$
$\Rightarrow \frac{1}{a^{2}}=\frac{17+12 \sqrt{2}}{289-288}$
$\Rightarrow \frac{1}{a^{2}}=17+12 \sqrt{2}$ ...........(2)
Subtracting (2) from (1), we get
$a^{2}-\frac{1}{a^{2}}=(17-12 \sqrt{2})-(17+12 \sqrt{2})$
$\Rightarrow a^{2}-\frac{1}{a^{2}}=17-12 \sqrt{2}-17-12 \sqrt{2}$
$\Rightarrow a^{2}-\frac{1}{a^{2}}=-24 \sqrt{2}$
Thus, the value of $a^{2}-\frac{1}{a^{2}}$ is $-24 \sqrt{2}$.
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