I need to show that (3 + i)^3 = 18 + 26i and use this to show that the angle AOC = 3AOB, where O, A, B, C are points in the plane given by O = (0, 0), A = (1, 0), B = (3, 1) and C = (18, 26).

This is what I have done:

Expand (3 + i)^3

(3 + i)^3 = (3 + i)(3 + i)(3 + i)

```
= 9 + 3i + 3i + i^2 * (3 + i)
= 9 + 6i - 1 * (3 + i)
= (8 + 6i) * (3 + i)
= 24 + 8i + 18i + 6i^2
= 24 + 26i - 6
= 18 + 26i
```

Find the magnitude of OB = 3 + i and OC = 18 + 26i to determine ->OB and ->OC

|OB| = sqrt(3^2 + 1^2) = sqrt(10)

|OC| = sqrt(18^2 + 26^2) = 10 * sqrt(10)

Therefore,

->OB = 3 + i = sqrt(10) * cis(AOB)

->OC = 18 + 26i = 10 * sqrt(10) * cis(AOC)

Cube ->OB and show to show AOC = 3AOB

->OB^3 = (3 + i)^3 = (sqrt(10))^3 * cis^3(AOB)

```
= 18 + 26i = 10 * sqrt(10) * cis^3(AOB)
```

We found that (3 + i)^3 = 18 + 26i and by cubing ->OB we find the magnitude is the same as ->OC. Since this is the case, this shows that the angle AOC = 3AOB.

Would this be the correct way to solve this question?