Perform the indicated operations and write the result in standard form. \[ (3+4 i)(3-i)-(2-i)(2+i) \] (A) $10+9 i$ (B) $8+9 i$ (C) $8+15 i$ (D) $18+9 i$

Step 1 ：First, multiply the complex numbers: \((3+4i)(3-i) = (3\times3 + 3\times(-i) + 4i\times3 + 4i\times(-i)) = (9 - 3i + 12i - 4i^2) = (9 - 4(-1) + 9i) = (13 + 9i)\)

Step 2 ：Next, multiply the other complex numbers: \((2-i)(2+i) = (2\times2 + 2\times i - i\times2 - i\times i) = (4 + 2i - 2i - i^2) = (4 - (-1)) = 5\)

Step 3 ：Finally, subtract the second result from the first result: \((13 + 9i) - (5) = (13 - 5) + 9i = 8 + 9i\)

Step 4 ：\(\boxed{8+9i}\)