Question

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

Answer 1

`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`