Question

Find all of the real and imaginary solutions of x^3+8=0

Answer 1

-2, 1 +/- i*sqrt(3)

(3 solutions)

Explanation:

Because the left side is in the form a^3+b^3, we can factor it as (a+b)(a^2-ab+b^2) (good formula to remember)

So we have

(x+2)(x^2-2x+4)

We can see one value already

x+2 = 0

x = -2

now we solve for

x^2-2x+4 = 0

We can use the quadratic formula, and see that the determinant is negative (2^2-4*1*4 < 0)

Therefore we have two imaginary solutions

(-(-2) +/- sqrt(2^2-4*1*4))/)2*1) = (2 +/- sqrt(-12))/2 = (2 +/- 2sqrt(-3))/2 = 1+/- sqrt(-3) = 1 +/- i*sqrt(3)

Answer 2

`x=-2, \:1+√(3)i,\: 1-√(3) i`

Explanation:

`x^3+8=0,\\x^3=-8,\\\mathrm{For\:}x\in \mathrm{R}, x=-2`

Real solutions (
`x \in \mathrm{R}`):
`x=-2`

However, recall that
`((-1\pm√(3)i)/(2))^3=1`.

Therefore:

`(-1\pm√(3)i)/(2)\cdot -2=-2`.

Thus, for
`x^3=-8,\: x\\otin \mathrm{R}`:

`x=1\pm√(3)i`.

Therefore, the real and nonreal solutions of
`x^3+8=0` are
`\fbox{$-2, \:1+√(3)i,\: 1-√(3)i$}`.