Proving G is Abelian from (a*b)^2=(a^2)*(b^2)

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Homework Statement


If G is a group such that (a*b)^2=(a^2)*(b^2) for all a,b in G, show that G must be abelian.


The Attempt at a Solution


First, I tried to expand the binomial (a*b)^2 and set it equal to (a^2)*(b^2). But then I didn't know where to go from there.
 
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You aren't trying very hard. A group is abelian if ab=ba for all a and b. (ab)^2=abab. (a^2)*(b^2)=aabb. Take it from there.
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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