find the value

Solution:

$a=3-2 \sqrt{2}$

$\Rightarrow a^{2}=(3-2 \sqrt{2})^{2}$

$\Rightarrow a^{2}=9+8-12 \sqrt{2}$

$\Rightarrow a^{2}=17-12 \sqrt{2}$ ..........(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 $-24 \sqrt{2}$.