6a2b2c2d2 + 2abcd2 can be factorised to 2abcd2*(3abc + 1)
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(CD - ab)^2 = (CD - ab)(CD - ab) = c^2 d^2 - 2abcd + a^2b^2 Try it with say a = 4, b = 3, c = 2 & d = 1: Then CD = 2 and ab = 12 so CD - ab = -10 and squared = 100 c^2 = 4 d^2 = 1 so c^2d^2 = 4 x 1 = 4 a^2 = 16 b^2 = 9 so a^2b^2 = 16 x 9 =144 2abcd = 48 giving 4 - 48 + 144 = 100. Shazam!
a(squared)+b(squared)=c(squared) The largest number is the c, and you can choose the a and b, since you need to find the a or the b. For example: You have the numbers 3, and 5 and you needed to find b. a(squared+b(squared)=c(squared) 3(squared)+b(squared)=5(squared) 9+b(squared)=25 -9 -9 b(squared)=16 Then take the square root b=4
it is asquared +b squared = c squared
a squared + b squared = c squared a=leg b=leg c=hypotenuse
The Pythagoream Thereom is a^2 + b^2 = c^2. Written out it is a squared plus b squared equals c squared.