If ab=1, how do you prove that a^2*b^2=1?

a ² b ² = ( ab ) ( ab ) = ( 1 ) ( 1 ) = 1

You could note that a^2=a*a
and use the fact that real number multiplication commutes

a^2b^2=a*a*b*b=a*b*a*b=(1)*(1)=1

it quite is a tricky question, considering if it quite is straight forward severe-college algebra there's a regulation of exponents that asserts: (ab)^n = a^n b^n for any integer n. so which you basically positioned n=-a million and you're carried out. It has not something to do with the "associative regulation". on the different hand you point out "summary algebra", which if actually real, ability you have not given us sufficient innovations to proceed.

ab = 1 → ab * ab = 1 * 1 → a^2 * b^2 = 1

Q.E.D.

ab = 1. a^2b^2 = a*a*b*b=a*b*a*b=1*1

Well there's a bunch of ways to do this, but the easiest would be to take the square root of a^2*b^2. Because that's just ab, and the square root of 1 is 1. So that equals 1.

You can also do substitution.
a= 1/b
Plug that into a^2*b^2
So you get (1/b)^2 * (b^2) = 1 because the b's cancel out, so you get 1 again.

Or you can factor out ab * ab from (a^2*b^2) that gives you 1*1=1.

a^2*b^2 = (ab)^2 = 1^2 = 1