Question

if x^3=1 then how many roots are possible in this equation and what is the roots?​

Answer 1

Answer:

One root is only possible and that is 1

Hope it helps you

Answer 2

Answer:

x^3 - 1 = 0 => (x-1)(x^2+x+1) = 0 => x-1=0 or x^2+x+1 = 0. So x = 1 is one of the roots.

To solve x^2+x+1=0, apply quadratic formula G is x {-b±(√(b^2–4ac)}/2a.

We get x = {-1 ± √(1^2 - 4(1)(1)}/2(1) = (-1±i√3)/2 which are imaginary.

The root (-1 + i √3)/2 is denoted by ω.

We find that {(-1 + i √3)/2}= = (-1 - i √3)/2^2 = ω^2

Hence the roots of x^3 = 1 are 1, ω, ω^2. These three roots of x^3=1 are called the cube roots of unity.

We also observe that 1+ ω+ω^2 = 0 and ω^3 = 1.