Question

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

Answer 1

i got 63

Answer 2

it could also be 15

Answer 3

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

lets try .. if we could simplify the equation further

and yes we can... -3a2 - b3 + 3c2 - 2b3
and since we could see b^3 twice terefore lets make it once

so now the equation becomes -3a^2 -3b^3 +3c^2
now even take the three common
we get 3 ( -a^2 - b^3 + c^2)
now using the value of a as a = 2
we get
3 ( -(2)^2 - b^3 + c^2)
i.e. 3( -4 - b^3 + c^2)
and now using b = -1
we get
3( -4 - (-1) + c^2)
now using c = 3
we get
3( -4 +1 + 9)
which is equals to 3(6) = 18...!