Question

3. What are the roots of the polynomial y = x³ - 8?

Answer 1

Step-by-step explanation:

x^3 is a perfect cube, 8 is a perfect cube, so we use difference of cubes.

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

Cube root of x^3 is x.

Cube root of 8 is 2

So

a=x

b= 2.

$(x - 2)( {x}^{2} - 2x + 4)$

Set these equations equal to zero

$x - 2 = 0$

$x = 2$

${x}^{2} - 2x + 4 = 0$

If we do the discriminant, we get a negative answer so we would have two imaginary solutions,

Thus the only real root is 2.

If you want imaginary solutions, apply the quadratic formula.

$1 + i \sqrt{ 3 }$

and

$1 - i \sqrt{3}$