Problem: If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is (i=−1​):
Answer Choices:
A. 2
B. −2−433​i​
C. 0
D. −433​i​
E. −2
Solution:
Let the numbers be x and y. Then x3+y3=(x+y)3−3xy(x+y) ∴x3+y3=13−3(1)(1)=−2
Consider the equation Z2−Z+1=0 with roots Z1​=21+i3​​,Z2​=21−i3​​.
The sum of these roots is 1 and their product is 1 , so that we may take for our numbers the numbers Z1​ and Z2​. Therefore Z1​3+Z2​3=(21+i3​​)3+(21−i3​​)3=81+3i3​+3i2⋅3+i3⋅33​+1−3i3​+3i2⋅3−i3⋅33​​=82−18​=−2