20 баллов двю.Срочноо. (-5+5i)в 3 степени=? (2+i)/(-3+i)=? (2+i)(-3+i)=?

Решите задачу:

$1)\quad (-5+5i)^3=(-5(1-i))^3=(-5)^3(1-i)^3=\\\\=-125(1-3i+3i^2-i^3)=-125(1-3i-3+i)=\\\\=-125(-2-2i)=-250(1+i)\\\\\star \; \; i^2=-1\; ,\; \; i^3=i^2\cdot i=-1\cdot i=-i\; \; \star \\\\2) \frac{2+i}{-3+i} = \frac{(2+i)(3+i)}{-(3-i)(3+i)} = \frac{6+5i+i^2}{9-i^2} = \frac{5+5i}{9+1} = \frac{5(1+i)}{10} = \frac{1+i}{2}= \frac{1}{2} +\frac{1}{2}i$

$3)\quad (2+i)(-3+i)=-6+2i-3i+i^2=-6-i-1=-7-i$