Express cos3Θ in terms of cosΘ

the answer is 4cos³Θ - 3cosΘ but I can't get it

Here's what I did:

cos3Θ + isin3Θ = (cosΘ + isinΘ)³ using De Moivre's theorem

(cosΘ + isinΘ)³ = cos³Θ + 3cos²ΘisinΘ - 3cosΘsin²Θ - isin³Θ using binomial theorem

I then equated the real parts: cos3Θ = cos³Θ - 3cosΘsin²Θ

and that's where I'm stuck. I have no idea how to get the correct answer. Anyone able to help? TIA!

You just need to sub sin²Θ in your answer with 1-cos²Θ

cos³Θ - 3cosΘsin²Θ = cos³Θ - 3cosΘ(1-cos²Θ)
= cos³Θ - 3cosΘ+3cos³Θ
= 4cos³Θ - 3cosΘ

I would do it with this:

cos(A+B) = cos(A)cos(B) - sin(A)sin(B)

cos(Θ + 2Θ) = cos(Θ)cos(2Θ) - sin(Θ)sin(2Θ)

use the same thing for cos(2Θ) = cos(Θ)cos(Θ) - sin(Θ)sin(Θ)
and use sin(2Θ) = sin(Θ)cos(Θ) + cos(Θ)sin(Θ) = 2sin(Θ)cos(Θ)

cos(3Θ) = cos(Θ)cos(2Θ) - sin(Θ)sin(2Θ) =
cos(Θ)[cos(Θ)cos(Θ) - sin(Θ)sin(Θ)] - sin(Θ)[2sin(Θ)cos(Θ)]
distribute:
cos^3(Θ) - cos(Θ)sin^2(Θ) -2cos(Θ)sin^2(Θ) =
cos^3(Θ) - 3cos(Θ)sin^2(Θ)

Use: sin^2(Θ) + cos^2(Θ) = 1
to replace sin^2(Θ) by 1 - cos^2(Θ)

cos^3(Θ) - 3cos(Θ)[1 - cos^2(Θ)] =
cos^3(Θ) - 3cos(Θ) + 3cos^3(Θ) =
4cos^3(Θ) - 3cos(Θ)

Don't go to all that trouble.

cos(3Θ)
= cos(Θ + 2Θ)
= cos(Θ)cos(2Θ) - sin(Θ)sin(2Θ)
= cos(Θ)(cos(Θ)^2 - sin(Θ)^2) - sin(Θ)(2cos(Θ)sin(Θ))
= cos(Θ)^3 - cos(Θ)sin(Θ)^2 - 2cos(Θ)sin(Θ)^2
= cos(Θ)^3 - 3cos(Θ)sin(Θ)^2
= cos(Θ)^3 - 3cos(Θ)(1 - cos(Θ)^2) [this is the step you're missing]
= cos(Θ)^3 - 3cos(Θ) + 3cos(Θ)^3
= 4cos(Θ)^3 - 3cos(Θ)

cos3Θ = cos³Θ - 3cosΘsin²Θ = cos3Θ = cos³Θ - 3cosΘ(1-cos²Θ) = 4cos³Θ - 3cosΘ