Question

What are the real and complex solutions of the polynomial equation? x^3-8=0. with imaginary numbers

Answer 1

Answer: The solutions are :
`2, -1 + √(3)i , -1 - √(3)i`

Our equation is
`x^(3) - 8 = 0`

and we know that:
`8 = 2^(3)`

then we can write our equation as:
`x^(3) - 2^(3) = 0`

And using the identity:
`a^(3) - b^(3) = (a-b)*(a^(2) +ab+ b^(2) )`

where a = x and b = 2, then our equation is:

`x^(3) - 2^(3) = (x-2)*(x^(2) +2x+ 2^(2) )= 0`

them, if x = 2 the first part is zero, so x = 2 is a solution, and now we need to see the second part in order to find the complex solutions.

`x^(2) + 2x + 4 = 0`

Here we need to use Bhaskara:

if
`ax^(2)+ bx + c=0` for a, b and c constants, then:

x =
`\frac{-b +-\sqrt{b^(2) - 4ac } }{2a}`

in our problem we have:

`x = (-2 +-√(4 - 16) )/(2) = -1 +-\sqrt{(4 - 16)/(4) } = -1 +- √(-3)  = -1 +-√(3) i`

So here we have two complex solutions:
`x = -1 + √(3)i` and
`x = -1 - √(3) i`.