# Can someone please hepl me with the rest of the problems in pre cal

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### 1 Answer

As per the policy of enotes, the educators do not complete the homework for the students, but guide them along the path. So I will solve some problems from each question, so that you can follow the lead and complete the assignment.

1. `(-2-2sqrt3 i)^3 = (-2)^3 (1+sqrt3 i)^3 = 8 [(1+sqrt3 i) (1+sqrt3 i)^2] `

`= 8[(1+sqrt3 i) (1+2sqrt3 i -3)] = 8[(1+sqrt3 i)(2sqrt3 i -2)] `

`= (8)(2)(sqrt3 i +1)(sqrt3i-1) = 16(-3-1) =16 xx-4 = -64`

We have simplified the terms, taken common terms outside the brackets, used the common expansions of (a+b)^2 and (a+b)(a-b).

5. `(2+3i)^6 = (2+3i)^(2+2+2)=([(2+3i)^2]^2)^2 = ((4+12i-9)^2)^2 `

`= ((12i-5)^2)^2 = (25-120i-144)^2 = (-120i-119)^2 = (-1)^2(120i+119)^2 `

`= -14400+ 14161+28560i = -239 + 28560i `

7. `(-27i)^(1/3) = (3X3X3 -i)^(1/3) =((3)^3)^(1/3) (-i)^(1/3) = 3 (-i)^(1/3) =3i = 0+3i `

10. `-i^(1/3) = (i X i X i)^(1/3)=(i^3)^(1/3) = i = 0+i`

-i can be written as the cube of i, since i X i = -1, i x i x i = (-1) i = -i

Pg. 390.

2. i

This is the root of equation of x^2= -1 or, x^2 + 1 = 0. Thus, i is algebraic.

You have already correctly solved 1 and 3 on pg. 390.

Hopefully you can get the ideas from these examples and solve the rest of the problems on your own.