Expanding binomials with binomial theorem / pascals triangle

Expand (5 √x +2)^3
(X is the only thing under the radical)


What I have:
Coefficients are 1, 3, 3, 1 because of Pascal's triangle
1 * 5√x^3 = 5√x^3
3 * 5√x^2 + 2 = 3(5√x^2 + 2)
3 * 5√x^1 + 2^2= 3(5√x + 4)
1 * 2^3 = 8

5√x^3 + 3(5√x^2 + 2) + 3(5√x + 4) + 8

I'm not sure how to simplify that? Besides distributing the threes, is there anything else you can do? And was I correct upto that point?

And for Expand (6-5i)^4
The coefficients are 1, 4, 6, 4, 1
So,
1* (6^4) = 1296
4* (6^3 - 5i) = 4(216 - 5i)
6* (6^2 - 5i^2)= 36 -5(-1) = 6(41)
4* (6 - 5i^3) = 4(6- 5(-i))
1* (1 - 5i^4) = 1-5 = -4

1296 + 4(216 - 5i) - 6(41) + 4(6- 5(-i)) - 1-5 = -4
Ok I'm not sure about the signs, I thought I saw somewhere that with a negative the signs alternated. But I'm not sure. Also, besides the obvious 6*41, how would I simplify, if I got upto that point correct?

Any help will be greatly appreciated.

(a+ b)³ = a³ +3a²b +3ab²+ b³ , a= (5 √x ) , b=2

= (5 √x )³+3 (5 √x )²(2)+3 (5 √x )(2)²+ 2³
=125(√x)³+6(25x) + 12*5√x +8
=125x√x+60√x+150x+8

(6-5i)^4
a=6
b= 5i
1(6)^4- 4(6)³(5i) +6(6)²(5i)² - 4(6)(5i)³ + 1(5i)^4
=1296 - 4320i +(-5400) - 3000(-i) +625(1)
=1296 - 4320i -5400 +3000i + 625
then simplify