Question

x= a + b, y = aa + bß and z= aß + ba, where a and ß are complex cube roots of unity, then what is the value of xyz?​

Answer 1

Answer:

Step-by-step explanation:

let complex cubes of unity be w and w²

Therefore w³=1

and 1+w²+w = 0 => w+w²= -1

x=a+b

y= wa+w²b

z= w²a+wb

Now y*z = (wa+w²b)*(w²a+wb)

= w³a² + w²ab + w⁴ab + w³b²

= a² + ab(w²+w) + b²

= a² - ab + b²

Now xyz = (a+b)(a²-ab+b²) = a³+b³