Question

Find all three solutions for the given equation.
$\sf \implies {x}^{3} = - 1$
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Answer 1

Answer :

x = -1 , (1 + √3i)/2 , (1 - √3i)/2

Solution :

Given : x³ = -1

To find : Solutions (Roots) of the given equation x³ = -1

We have ,

=> x³ = -1

=> x³ + 1 = 0

=> x³ + 1³ = 0

=> (x + 1)(x² - x•1 + 1²) = 0

=> (x + 1)(x² - x + 1) = 0

Here ,

Two cases arises ;

• x + 1 = 0 OR

• x² - x + 1 = 0

•Case1 :

If x + 1 = 0 , then x = -1

• Case2 :

If x² - x + 1 = 0 , then

This is a quadratic equation in x .

Now ,

Comparing the equation x² - x + 1 = 0 with the general quadratic equation ax² + bx + c = 0 , we have ;

a = 1

b = -1

c = 1

Now ,

The discriminant will be given as ;

=> D = b² - 4ac

=> D = (-1)² - 4•1•1

=> D = 1 - 4

=> D = -3

=> D < 0 , thus the equation x² - x + 1 = 0 will have complex conjugate pair of roots .

Now ,

The roots will be given as ;

x = ( -b ± √D)/2a

x = [-(-1) ± √(-3)]/2•1