How can I simplify this expression using factoring?

AI Thread Summary
To simplify the expression 2a^2*b^2*c^2 + (a+c)^2*(a+b)^2*(b+c)^2 - a^2*c^2(a+c)^2 - (a+b)^2*a^2*b^2 - (b+c)^2*b^2*c^2 = 2abc(a+b+c)^3, it is recommended to expand and group the terms strategically. Using factoring techniques, such as recognizing patterns like (a-b)^2 = (a+b)(a-b), can aid in simplification. Breaking down the expression into manageable parts and working through them individually may reveal a clearer path to the solution. Drawing lines to separate terms can help visualize the problem better. Persistence and multiple attempts are key to tackling complex expressions.
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Homework Statement


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2a^2*b^2*c^2 + (a+c)^2*(a+b)^2*(b+c)^2 - a^2*c^2(a+c)^2 - (a+b)^2 * a^2 * b^2 - (b+c)^2*b^2*c^2 = 2abc(a+b+c)^3


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Well I actually tried to expand all of the exponential terms but that ended up being a total mess and wasn't even remotely obvious... My professor said using something like (a-b)^2 = (a+b)(a-b) would help, but I don't quite see how that would help.. can anyone give me a hint?
 
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ya only tried once? I find that I need to work through complicated expressions multiple times. Don't give up keep after it. You may want to show us some of your work. If you do pleas go to the tutorials forum and read the latex thread.
 
Try it in parts. When I solve problems that look like a mess I'll draw lines down a paper and group it by addition/subtractions. That way you're just dealing with each part individually.

<br /> \underbrace{2a^2*b^2*c^2}_{First} + \underbrace{(a+c)^2*(a+b)^2*(b+c)^2}_{Second} - \underbrace{a^2*c^2(a+c)^2}_{Third} - \underbrace{(a+b)^2 * a^2 * b^2}_{Fourth} - \underbrace{(b+c)^2*b^2*c^2}_{Fifth} &amp;=&amp; \underbrace{2abc(a+b+c)^3}_{Sixth}<br />

Now that you've grouped off your terms. Expand these out, simplify. The solution may become obvious (or at least easier) to find. When you're done, you may be left with something as simple as a quadratic, but don't quote me on that. I've not done the problem myself.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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