$\frac{3+\mathrm{i}}{(3-\mathrm{i})^{2}}+\frac{3-\mathrm{i}}{(3+\mathrm{i})^{2}}=\frac{9}{25}$
\(\boxed{\frac{9}{50}}\)
Step 1 :\(\frac{3+i}{(3-i)^2} + \frac{3-i}{(3+i)^2}\)
Step 2 :Multiply the first term by \(\frac{3+i}{3+i}\) and the second term by \(\frac{3-i}{3-i}\)
Step 3 :\(\frac{(3+i)^2}{(3-i)^2(3+i)} + \frac{(3-i)^2}{(3+i)^2(3-i)}\)
Step 4 :Combine the terms with a common denominator
Step 5 :\(\frac{(3+i)^2(3-i) + (3-i)^2(3+i)}{(3-i)^2(3+i)^2}\)
Step 6 :Expand the numerator
Step 7 :\(\frac{9+6i+i^2+9-6i-i^2}{(3-i)^2(3+i)^2}\)
Step 8 :Simplify the numerator using \(i^2=-1\)
Step 9 :\(\frac{18}{(3-i)^2(3+i)^2}\)
Step 10 :Expand the denominator
Step 11 :\(\frac{18}{(9-6i+1)(9+6i+1)}\)
Step 12 :Simplify the denominator
Step 13 :\(\frac{18}{100}\)
Step 14 :Reduce the fraction
Step 15 :\(\boxed{\frac{9}{50}}\)