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Hey man, I think I sent you some messages (sorry if I double posted)

Dammit got in quicker than me. I blame my laggy computer, which made my browser freeze for 5 minutes :(

Question 13: x(a-b)+ y(b-a) = x(a-b) - y(a-b) , by taking out the negative of (b-a) = (x-y) (a-b) , using the grouping method So the grouping method is basically, as follows: If you're given a(c+d) + b(c+d) , it can be "grouped" (factorised is a better...

wassup is da ceiling for me lol...hbu? :L

lol hows it going? havent been on in ages

Nice new dp brah (Y)

Sigh...Ben...

You can write a^4 + 4b^4 as: a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 = (a^2 + 2b^2)^2 - (2ab)^2 after completing the square and rearranging as a difference of two squares. now using the difference of two squares, we can factorise it to: (a^2 + 2b^2 + 2ab) (a^2 + 2b^2 - 2ab) rearranging the terms...

yeah, im from tech. Pleased to meet u =P