Ifa+b+c=0,show that,(a+bomega+comega^2)^3+(a+bomega^2+comega)^3=27abc?

1 Answer
May 29, 2018

a + b + c = 0

LHS = (a + bω + cω²)³ + (a + bω² + cω)³

If we consider A = (a + bω + cω²) and B = (a + bω² + cω)

Adding the two.

color(magenta)(A + B) = 2a + b(omega+omega^2) + c(omega^2 +omega)
=> 2a-b-c = 3a-a-b-c = 3a-0 = color(magenta)(3a) "as " [ω² + ω + 1 = 0]

Multiplying the two

color(magenta)(AB) = (a + b + c)² - 3(ab + bc + ca) = color(magenta)(- 3(ab + bc + ca)

As we know, (A+B)^3 = A^3+B^3 +3AB(A+B)

P => A^3+B^3 = (A + B)³ - 3AB(A + B)
= 27a³ + 27a(ab + bc + ca)
= 27a(a² + ab + ac + bc) = 27a(a + b)(a + c) = 27a(-c)(-b) = 27abc