SOLUTION: 1.) Perform the indicated operations: a.) b^3 - b^2 + 3b - (b - 3) + (3 - 2b^2 + 3b^3)= b.) -3(3 - b)(3 + b) = c.) (c - 3)^2= d.) (a - b)(a^2 + ab + b^2)= e.) 6c^2 -

Algebra ->  Functions -> SOLUTION: 1.) Perform the indicated operations: a.) b^3 - b^2 + 3b - (b - 3) + (3 - 2b^2 + 3b^3)= b.) -3(3 - b)(3 + b) = c.) (c - 3)^2= d.) (a - b)(a^2 + ab + b^2)= e.) 6c^2 -      Log On


   

Question 80035: 1.) Perform the indicated operations:
a.) b^3 - b^2 + 3b - (b - 3) + (3 - 2b^2 + 3b^3)=
b.) -3(3 - b)(3 + b) =
c.) (c - 3)^2=
d.) (a - b)(a^2 + ab + b^2)=
e.) 6c^2 - 12c +3 / -3c =
Answer by sofiyacherni(99) About Me  (Show Source):
You can put this solution on YOUR website!
1.) Perform the indicated operations:
a.) b^3 - b^2 + 3b - (b - 3) + (3 - 2b^2 + 3b^3)= b^3-b^2+3b-b+3+3-2b^2+3b^3=4b^3-3b^2+2b+6
b.) -3(3 - b)(3 + b) = -3(9-3b+3b-b^2)= -3(9-b^2)=-27+3b^2
c.) (c - 3)^2= c^2-6c+9
d.) (a - b)(a^2 + ab + b^2)= a^3+a^2b+ab^2-a^2b-ab^2-b^3=a^3-b^3
e.) 6c^2 - 12c +3 / -3c = 3(2c^2-4c+1)/-3c = (2c^2-4c+1)/-c you can distribute the minus sign also to get
(-2c^2+4c-1)/c