1/2 + (i times square root of 3)/2=a cube root of -1?

Question

how can you show that's true? i have no idea where to begin so please show all steps. thanks so much!

Answer 1

(1/2 + i(1/2)√3)^3 =? - 1

(1/2)^3 + i(3)(1/2)^3(√3) + i^2(3)(1/2)^3(3) + i^3(1/2)^3(3√3) =? - 1

(1/8)(1 + i(3)(√3) + i^2(3)(3) + i^3(3√3)) =? - 1

(1/8)(1 + i(3)(√3) - (3)(3) - i(3√3)) =? - 1

(1/8)(1 - 9) =? - 1

(1/8)(- 8) =? - 1

-1 = -1

Answer 2

Just multiply it out. You might use the fact that (A+B)^3 = A^3 + 3(A^2)B + 3A(B^2) + B^3

In this problem, A = 1/2, and B = (i times square root of 3)/2

Answer 3

Gawd, so long ago....

There are google sites with solutions to such things.

What is the square root of -1? i am!
OK...

Try reversing the equation? Cube both sides, so that :

-1 = (1/2 + iz^)*3 ^ z = (i times square root of 3)/2
= 1/8 + iz/2 + .... gaaahhhh...*runs Away*******

Answer 4

12

Answer 5

Everybody above used a different method. I would do it as a problem in complex numbers.

-1 = i^2. So cube root of (-1) =(-1)^1/3 = (i^2)^1/3 = i^(2/3)

i = 0 +i = cos(pi/2) + i sin(pi/2)
i^(2/3) = [cos(pi/2) +isin(pi/2)]^2/3
Using De Moivre's theorem ( easy to prove by using exponential expressions for sin and cos functions),

we have [cos(a) + isin(a)]^n = [cos(na)+isin(na)]

So i^(2/3) becomes [cos(2/3)(pi/2) +i sin(2/3)(pi/2)]
=[cos(pi/3) +i sin(pi/3)]
=[ 1/2 + i (sqrt3)/2]

Answer 6

Multiply it out ...
x = (1/2)(1 + i*sqrt(3))
x^2 = (1/4)(1 + 2i*sqrt(3) - 3) = (1/2)(-1 + i*sqrt(3))
x^3 = (1/4)(1 + i*sqrt(3))(-1 + i*sqrt(3))
= (1/4)(-1 -3)
= -1

Answer 7

go to math.com it might help you

Answer 8

answer HOLY ****