Verify if the number (1+i*sqrt3)^3 is real .

Expert Answers
hala718 eNotes educator| Certified Educator

(1+i*sqrt3)^3

Let us rewrite:

==> (1+ i*sqrt3)^3 = (1+i*sqrt3)^2 *(1+isqrt3)

                               = (1+ 2i*sqrt3 -3) (1+i*sqrt3)

                             = 1 + i*sqrt3 + 2isqrt3 -2*3 - 3isqrt3

                            = 1+3sqrt3 - 6 - 3sqrt3

                             = -5 + 0

Then the result is a real number .

thewriter | Student

To verify that (1+i*sqrt 3)^3 is real find the value of (1+i*sqrt 3)^3 and check if i is present. In case i has been eliminated it is real.

Now (a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3

So (1+i*sqrt 3)^3=1-3*3+3*i*(sqrt3)+3*(sqrt3)*-1*i

=1-9=-8

Therefore as there is no i in the result, (1+i*sqrt 3)^3 is real.

giorgiana1976 | Student

First, we have to calculate the number and we'll do it by expanding the cube.

We'll apply the formula

(a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3

(1+i*sqrt3)^3=1^3+3*1^2*sqrt3+3*1*(sqrt3)^2+(sqrt3)^3

We'll combine and eliminate like terms:

1+3(sqrt3)*i-9-3(sqrt3)*i=-8

The value of the number (1+i*sqrt3)^3 = -8, is a real number.