other formula?

Question

is
a^(2)-b^(2)
=(a-b)^(2)+2b(a-b)
true?

Answer 1

True

Answer 2

This doesn't look like a formula to me.

I call it the 'brute force' way. Usually used during tests and exams when we know the destinations, i.e. the target expression.

E.g. in the working half way you arrive at a^2-b^2, but you know the answer is or of the form of k(a-b)^2 + ...

Just write first term down, you will have
k(a-b)^2, which gives you ka^2 -2kab + kb^2,
but you need only 1 a^2, k=1, so you get a^2-2ab+b^2
but you don't need -2ab while you need a (-1)*b^2, so remove the -2ab by adding 2ab and add -2b^2 so that b^2 can become -b^2, so you get
(a^2-2ab+b^2) + 2ab - 2b^2
= (a-b)^2 +2b(a-b)
The above mental process only takes few seconds, and can be skipped, i.e. cheating, if you are confident enough and the destination is exactly given.

Answer 3

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

= a^2 - b^2

Hence true.

Answer 4

a^2 - b^2 = (a-b)^2 + 2b(a-b)
= a^2 + b^2 - 2ab + 2ab - 2b^2
= a^2 + b^2 - 2b^2
= a^2 - b^2
Where did you take maths?

Answer 5

yes,it is
(a-b)^2+2b(a-b)=a^2+b^2-2ab+2ab-2b^2=a^2-b^2

Answer 6

Yes it is. Just do the maths

Answer 7

yes that is equal i visualize it in my brain cause that is simple. I don't visual other than that simply.

Answer 8

yes it is

Answer 9

yes and you know that.