To solve for the value of
xy+yz+zx where
x,y, and
z are the cube roots of unity, let's use the properties of the cube roots of unity. The cube roots of unity are the solutions to the equation:
z3=1These solutions are:
1 (which is the real root)
and the two complex roots:
ω=e2πi∕3ω2=e4πi∕3These roots satisfy the equation:
x3=1y3=1z3=1One important property of the cube roots of unity is that their sum is zero:
x+y+z=0Additionally, these roots satisfy the following relationship:
xyz=1 We need to find the value of
xy+yz+zx. Using the identity for the sum of the products of the roots taken two at a time for a polynomial equation, we have:
x3−1=0The polynomial can be expressed as:
(z−1)(z−ω)(z−ω2)=0Expanding this product, we get:
z3−(1+ω+ω2)z2+(ω+ω2+ω⋅ω2)z−ω⋅ω2=0Since
1+ω+ω2=0 (as they sum to zero), and since
ω⋅ω2=ω3=1 because they are roots of unity, we can simplify the expanded polynomial to:
z3−z=0Therefore, the expression simplifies to:
xy+yz+zx=xixj=0 (since the sum of the roots is zero as well)