Question Detail

Let z = 2+i, then (2-i)³ - (2+i)³ = (conjugate of z)³ - z³. For complex numbers, (a-bi)³ - (a+bi)³. Compute (2+i)³ = 8 + 12i + 6i² + i³ = 8+12i-6-i = 2+11i. Then (2-i)³ = 8 -12i +6i² -i³ = 8-12i-6+i = 2-11i. Difference = (2-11i)-(2+11i) = -22i. Not matching. Maybe (2-i)⁴ - (2+i)⁴? (2+i)² = 3+4i, (2+i)⁴ = (3+4i)² = 9+24i+16i² = -7+24i. (2-i)² = 3-4i, (2-i)⁴ = (3-4i)² = 9-24i+16i² = -7-24i. Difference = (-7-24i)-(-7+24i) = -48i. Not matching. For (2-i)³ - (2+i)³, using formula a³-b³ = (a-b)(a²+ab+b²), with a=2-i, b=2+i, a-b=-2i, a²+b² = (3-4i)+(3+4i)=6, ab=(2-i)(2+i)=5, so a²+ab+b² = 6+5=11, product = (-2i)×11 = -22i. Still -22i. Option C is -242i = 11×(-22i), so maybe they have a factor of 11 different.