Can you solve this algebraic sum

Using formula find the value of 8-9a+27a2-27a3, when a= 2
{NOTE THAT THE 2 AFTER SECOND 'a' is square AND THE 3 AFTER THIRD 'a' is cube }

essentially you just replace all of the a's with a 2
(also to show powers use ^, ex. 2 squared would be 2^2)

so the equation would look like this:
8 - 9(2) + 27(2^2) - 27(2^3)
then you simplify the exponents in the parenthesis
8 - 9(2) + 27(4) - 27(8)
then you mulitply the parenthesis
8 - 18 + 108 - 216
then its just simple adding and subtracting
so the answer is: -118

I got =
BIDMAS (brackets, indesies, devision, multiplication, addition and subtraction) in order
=8-9a+27a2-27a3
=8-18+27x4-27x8
=8-18+108-216
=8-126-216
=-334

Just to check. We have

8 - 9a + 27 a^2 - 27 a^3

Well, when a = 2 then

a^2 = 4 and a^3 = 8

So

8 - 9(2) + 27 (4) - 27 (8) = 8 - 18 + 108 - 216 = - 118

8 - 9a + 27a^2 - 27a^3
= 8 - 9(2) + 27(2)^2 - 27(2)^3
= 8 - 18 + 27(4) - 27(8)
= -10 + 108 - 216
= -118

8 - 18 + 27 x 4 - 27 x 8

- 10 + 108 - 216

108 - 226

- 118

8-9*2+27*2^2-27*2^3 = 8-18+108-216

The answer is -118