Question

x³– 1

solve it............​

Answer 1

$\huge\underline{ \mathbb\red{❥A} \green{n}\mathbb\blue{S} \purple{w} \mathbb \orange{E}\pink{r}}\:$

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.

$\huge</p><p>\underline{\red{Hope\:It\:Helps\:}}</p><p>$

Answer 2

using the identity

${a}^{2} - {b}^{2} = (a - b) \times {a}^{2} + ab + {a}^{2}$

$so ,$

$( {x}^{3} ) - 1 = (x - 1) \times ( {x }^{2} + x + 1)$