(2a2 (b+c)2n - 1/2) x 2/2 = (4a2 (b+c)2n - 1) x 1/2
Let’s keep the 1/2 aside for now.
We can write 4a2 (b+c)2n as 22 a2 ((b+c)n )2 and 1 as 12
So (4a2 (b+c)2n - 1) = ( 2a(b+c)n )2 - 12
Comparing with x2 - y2 we have x = 2a(b+c)n and y = 1
With with x2 - y2 =(x+y) (x-y), we get:
(2a(b+c)n)2 - 12 = (2a(b+c)n + 1) (2a(b+c)n - 1)
We can then cancel the common (2a(b+c)n - 1) term from the numerators on both sides…
with the assumption that 2a(b+c)n - 1 ≠ 0. If it’s zero, then we introduce the issue of 0 x something = 0 x something.
0 x 1 = 0 x 2 = 0
We can’t cancel the common zero and say that 1 = 2. So 0/0 or 1/0 is an issue, but with the assumption that it isn’t 0, we can cancel it out.
I multiplied and dividing by 2 because in effect it changes nothing to the result as 2/2 = 1. But it helps to rearrange and simplify stuff.
(2a2 (b+c)2n - 1/2) x 2/2 = (4a2 (b+c)2n - 1) x 1/2
Let’s keep the 1/2 aside for now.
We can write 4a2 (b+c)2n as 22 a2 ((b+c)n )2 and 1 as 12
So (4a2 (b+c)2n - 1) = ( 2a(b+c)n )2 - 12
Comparing with x2 - y2 we have x = 2a(b+c)n and y = 1
With with x2 - y2 =(x+y) (x-y), we get:
(2a(b+c)n)2 - 12 = (2a(b+c)n + 1) (2a(b+c)n - 1)
We can then cancel the common (2a(b+c)n - 1) term from the numerators on both sides…
with the assumption that 2a(b+c)n - 1 ≠ 0. If it’s zero, then we introduce the issue of 0 x something = 0 x something.
0 x 1 = 0 x 2 = 0
We can’t cancel the common zero and say that 1 = 2. So 0/0 or 1/0 is an issue, but with the assumption that it isn’t 0, we can cancel it out.
I multiplied and dividing by 2 because in effect it changes nothing to the result as 2/2 = 1. But it helps to rearrange and simplify stuff.
Ah gotcha. Now it’s all clear.