show that 1+ i√3 is a cube root of √-8?

Question

could you please show the steps

Answer 1

If it is a cube root, then cubing it should get back to -8

In other words:
(1 + i√3)^3 should equal -8.

Expand that out:
(1 + i√3)(1 + i√3)(1 + i√3)

FOIL on the first two parentheses:
(1 + 2i√3 + i²(√3)²)(1 + i√3)

Because i² = -1 and (√3)² = 3 we can simplify to:
(1 + 2i√3 - 3)(1 + i√3)

Simplify 1 - 3 :
(-2 + 2i√3)(1 + i√3)

Now foil these two:
(-2 - 2i√3 + 2i√3 + 2(i²)(√3)² )

The terms in the middle cancel:
(-2 + 2(i²)(√3)² )

And we can simplify the second term:
(-2 + 2(-1)(3) )
(-2 + -6)
= -8

QED

Answer 2

First let's start with the definition:

The cube root of a number x is the number, or numbers, multiplied by itself three times that equals x.
So, before I tell you the method, we should simplify √-8.
First let's write √-8 as i√8. This works because i=√-1, and this definition prooves it further:

For any numbers (which are not both negative),

√x*√y=√x*y.

So we now have it in the form: i√8
We're not done yet, though. √8 can be simplified. Using the rule I just stated, we can express √8 as:

√2*√2*√2

Now, since the square root of a number squared is that number, we can take two of the square root two's (since they're squared) and set them equal to two. The last √2 can't be simplified. Therefore we have:

2i√2.

Now all that has to be done to prove that this is a cubic root of (1+i√3) is to multiply it by itself three times.
Here we use the FOIL method (First, Outer, Inner, Last):

(1+i√3)(1+i√3) = 1²+i√3+i√3+3i²

Now let's simplify and combine the like terms:

1+2i√3+3(-1) = 1+2i√3-3 = -2+2i√3.

Now that we've squared the cubic root, I need to multiply this by the cubic root again and if I get 2i√2 (√-8 simplified), than this is one of the cubic roots.

(-2+2i√3)(1+i√3) = -2-2i√3+2i√3+2i²*3

The middle terms cancel out:

-2+2(-1)(3) = -2+(-2)(3) = -2-6 = -8

So it turns out that (1+i√3) isn't a cube root of √-8 (2i√2), but it's the cube root of -8.
There you go. I'm glad I could help. :)