Find all cubic roots of -8/ verify one complex root?

Question

So I figured out that the 3 cubic roots are -2, 1+i(square root of 3) and 1-i(square root of 3)

So how do i verify that 1+i(square root of 3) multiplied by 3 is -8?
Is it?

Answer 1

It sounds like you already have the method of determining the cubic roots of -8. You start with x^3 = -8 and solve for x. This becomes x^3 + 8 = 0. You know that -2 is a real root, so factor out (x + 2) from x^3 + 8. This leaves x² - 2x + 4 = 0. Then use the quadratic formula to solve.

To check your answers, you don't multiply each one by 3... you cube it.

-2 is obvious:
-2 x -2 x -2 = -8

Here's how you would do it for one of the complex roots:
(1 + i√3)(1 + i√3)(1 + i√3)
= (1 + 2i√3 - 3)(1 + i√3)
= (-2 + 2i√3)(1 + i√3)
= -2 + 2i√3 - 2i√3 + 2(i²)3
= -2 + 6(i²)
= -2 + -6
= -8

There you go... and you could do the same with the 3rd root, if you liked.

Answer 2

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Finding the cubic roots of -8 is equivalent to finding the roots of the equation
x^3 +8 =0

we know that x =-2 is a root. to find the other 2 roots

x^3 +8 = (x +2)(ax^2 +bx +c), expand this to get

ax^3 +2ax^2 +bx^2 +2bx +cx +2c =
ax^3 +(b +2a)x^2 +(c +2b)x +2c = x^3 +8

by identification

a =1
b+2a =0 ---> b =-2a =-2
c +2b =0
2c = 8 ---> c =4

ax^2 +bx +c = x^2 -2x +4

x^2 -2x +4 =0
D = 4 -16 = -12 = 12 i^2

x1 = (2 + 2i √ 3)/2 = 1 +i√ 3
x2 = 1 -i√ 3

the cubic roots of 8 are

-2
1 - i√ 3
1 + i√ 3

To verify that the 3 roots are roots of -8, expand the following
(x +2) ( x -1 + i√ 3)( x -1 - i√ 3)=0 , you find
x^3 +8 =0 or x^3 =-8 (hence the 3 roots are correct)

Answer 3

i = ²√(-1)

so 1 + i√(3) = 1 + √(-1*3) = 1 + √(-3)
but the √3 or √(-3) when added or subtracted from one does not = 2 ... therefore it is NOT a cubic root of -8

Answer 4

I know for +8

The three cube roots of 8 are:
a) {2 + i0}
b) {2*sin(2pi/n) + i2*cos(2pi/n)} (n = 3)
c) {-1.732 -i}

Find the point in the complex plane {sqrt(x)+ i0} where
sqrt(x) is the real positive root.
Draw a circle in the complex plane through the point,
centre on the origo to do.
Divide the circle into n equal segments to do.
Where the segments meet and intersect the circle are
the n-th roots of (x+ i0).


Sanity check to do:

(a+ib)^2 = {a^2 -b + i2ab) * (a+ib)

{-1.732 -i} * {-1.732 -i} * {-1.732 -i} = 8

Answer 5

Cubic root is - 2

NO complex roots for cubic root of - 8, unless you do sqrt (-8)