Question

If w is a complex cube root of unity, than prove the following (w²+w-1)³=-8

Answer 1

Before, first let me find the cube roots of unity.

x³ = 1

x³ - 1 = 0

(x - 1)(x² + x + 1) = 0

From this, we get,

x = 1

And,

x² + x + 1 = 0.

The roots of this quadratic equation are complex numbers and are usually considered as ω (omega) and ω² (omega squared).

From the quadratic equation, let a = b = c = 1.

ω² + ω = - b / a = - 1

Now let's prove what we're given.

LHS

=> (ω² + ω - 1)³

=> (- 1 - 1)³

=> (- 2)³

=> - 8

=> RHS

Hence Proved!