Question

From the sum of a2−2ab+b2 and 2a2+2ab+b2 subtract the sum of a2−b2 and a2+ab+3b2

Every 2 in this problem is meant to be a power of 2

Answer 1

Answer: So the other guy's work is correct but he messed up on the answer. The answer is
`-a^2+ab\\`.

Answer 2

`a^2-ab`

Explanation:

We need to find the sum of
`a^2-2ab+b^2` and
`2a^2+2ab+b^2` first.

Adding
`a^2-2ab+b^2` +
`2a^2+2ab+b^2`

Combining like terms, we get

`a^2+2a^2=3a^2`

-2ab+2ab = 0

`b^2+b^2=2b^2`

Therefore,

`a^2-2ab+b^2 +2a^2+2ab+b^2=3a^2+2b^2`.

Now, we need to find the sum of
`a^2-b^2 \ \ and \ \ a^2+ab+3b^2.`.

Adding
`a^2−b^2+a^2+ab+3b^2`.

Combining like terms, we get

`a^2+a^2=2a^2`

`-b^2+3b^2=2b^2`.

Therefore,

`a^2−b^2+a^2+ab+3b^2=2a^2+ab+2b^2.`

Now, subtracting

`3a^2+2b^2 -(2a^2+ab+2b^2)`.

Distributing minus sign over second parenthesis, we get

`3a^2+2b^2-2a^2-ab-2b^2`.

Combining like terms,

`3a^2-2a^2=a^2`

`2b^2-2b^2=0`

Therefore,

`3a^2+2b^2-2a^2-ab-2b^2=a^2-ab`.

Therefore, the difference of the sum of
`a^2-2ab+b^2` +
`2a^2+2ab+b^2` and
`a^2-b^2+a^2+ab+3b^2.` is
`a^2-ab.`