Question

What will three roots of x ³ -1 = 0


quality answer required !!

Answer 1

Given Equation is x^3 - 1 = 0

= > x^3 - 1^3 = 0

We know that a^3 - b^3 = (a - b)(a^2 + ab + b^2).

Now,

= > (x - 1)(x^2 + x + 1) = 0

(1) 

x - 1 = 0

x = 1.


(2) 

x^2 + x + 1 = 0

On comparing with ax^2 + bx + c, we get a = 1, b = 1, c = 1.

(i) 

$= \ \textgreater \ x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-(1) + \sqrt{(1)^2 - 4 * 1 * 1} }{2(1)}$

$= \ \textgreater \ \frac{-1 + \sqrt{1 - 1 * 1 * 4} }{2}$

$= \ \textgreater \ \frac{-1 + \sqrt{3}i }{2}$

$= \ \textgreater \ - \frac{1}{2} + \frac{ \sqrt{3} }{2} i$



(ii)

$= \ \textgreater \ x = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-(1) - \sqrt{(1)^2 - 4 * 1 * 1} }{2(1)}$

$= \ \textgreater \ \frac{-1 - \sqrt{3}i }{2}$

$= \ \textgreater \ \frac{-1}{2}- \frac{ \sqrt{3} }{2} i$


Therefore the required solutions are:

$= \ \textgreater \ x = 1, -\frac{1}{2} + \frac{ \sqrt{3} }{2} i, -\frac{1}{2} - \frac{ \sqrt{3} }{2} i$




Hope this helps!

Answer 2

Given Equation is x^3 - 1 = 0

= > x^3 - 1^3 = 0

We know that a^3 - b^3 = (a - b)(a^2 + ab + b^2).

Now,

= > (x - 1)(x^2 + x + 1) = 0

(1) 

x - 1 = 0

x = 1.


(2) 

x^2 + x + 1 = 0

On comparing with ax^2 + bx + c, we get a = 1, b = 1, c = 1.

(i) 

= \ \textgreater \ x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}= > x=2a−b+b2−4ac​​ 

= \ \textgreater \ \frac{-(1) + \sqrt{(1)^2 - 4 * 1 * 1} }{2(1)}= > 2(1)−(1)+(1)2−4∗1∗1​​ 

= \ \textgreater \ \frac{-1 + \sqrt{1 - 1 * 1 * 4} }{2}= > 2−1+1−1∗1∗4​​ 

= \ \textgreater \ \frac{-1 + \sqrt{3}i }{2}= > 2−1+3​i​ 

= \ \textgreater \ - \frac{1}{2} + \frac{ \sqrt{3} }{2} i= > −21​+23​​i 



(ii)

= \ \textgreater \ x = \frac{-b - \sqrt{b^2 - 4ac} }{2a}= > x=2a−b−b2−4ac​​ 

= \ \textgreater \ \frac{-(1) - \sqrt{(1)^2 - 4 * 1 * 1} }{2(1)}= > 2(1)−(1)−(1)2−4∗1∗1​​ 

= \ \textgreater \ \frac{-1 - \sqrt{3}i }{2}= > 2−1−3​i​ 

= \ \textgreater \ \frac{-1}{2}- \frac{ \sqrt{3} }{2} i= > 2−1​−23​​i 


Therefore the required solutions are:

= \ \textgreater \ x = 1, -\frac{1}{2} + \frac{ \sqrt{3} }{2} i, -\frac{1}{2} - \frac{ \sqrt{3} }{2} i= > x=1,−21​+23​​i,−21​−23​​i .