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:

$a^2-ab$

Step-by-step 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 $a^2-ab.$