Question

(b^2+2b)^2-b^2 (b-1)(b+1)+2b (3-2b)

Answer 1

(b^2+2b)^2-b^2 (b-1)(b+1)+2b (3-2b)=b^4+4b^2+4b^2-b^2(b^2-1)+6b-4b^2=b^4-1+6b-4b^2
(Вроде бы так) 

Answer 2

Let's simplify the expression step by step.

First, we can expand the square of the binomial (b^2 + 2b)^2 using the formula:

(a + b)^2 = a^2 + 2ab + b^2

So, we get:

(b^2 + 2b)^2 = b^4 + 4b^3 + 4b^2

Next, we can expand the product b^2 (b-1)(b+1) using the difference of squares formula:

(a + b)(a - b) = a^2 - b^2

So, we get:

b^2 (b-1)(b+1) = b^2 (b^2 - 1) = b^4 - b^2

Now, let's simplify the last term 2b (3-2b):

2b (3-2b) = 6b - 4b^2

Putting all the terms together, we get:

(b^2 + 2b)^2 - b^2 (b-1)(b+1) + 2b (3-2b) = (b^4 + 4b^3 + 4b^2) - (b^4 - b^2) + (6b - 4b^2) = 4b^3 + 9b^2 + 6b

Therefore, the simplified expression is 4b^3 + 9b^2 + 6b.