Find -3a2 - b3 + 3c2 - 2b3 if a = 2, b = -1, and c = 3

If you would like to solve - 3 * a^2 - b^3 + 3 * c^2 - 2 * b^3, if a = 2, b = -1, c = 3, you can calculate this using the following steps:

a = 2, b = -1, c = 3
- 3 * a^2 - b^3 + 3 * c^2 - 2 * b^3 = - 3 * 2^2 - (-1)^3 + 3 * 3^2 - 2 * (-1)^3 = - 3 * 4 - (-1) + 3 * 9 - 2 * (-1) = - 12 + 1 + 27 + 2 = 18

The correct result would be 18.

The first thing you want to do, is to substitute the  given to the equation. It will look like this:
-3(2)2 + 3(3)2 -2(1)3. 
Remember the MDAS method (Multiply, Division, Addition and Subtract.) So, you need to multiply all the numbers we have substitute in the equation.
-12 + 18 - 6. 
Then, remember to add comes first and subtraction comes last. So, we need to add -12 and +18 that will lead to this:
+6 - 6
Since -12 is negative and +18 is postive, so if we’ll add them up it’s +6. It is positive since +18 is bigger than -12. The answer is zero (0.)

/-/////////looks tough ‍♂️