$\sqrt{3 - \sqrt{5}}\cdot\left(3 + \sqrt{5}\right) \cdot \left(\sqrt{10} - \sqrt{2}\right)$

Can you please solve this? I've been trying to do that two hours! The result is 8.

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Square it: you should get 64. –  Robin Chapman Oct 3 '10 at 16:43
you have not understood Robins advice or followed it wrong. It is the best way to evaluate this number. –  anon Oct 3 '10 at 17:07
This is similar to your last question. You'll get more from this site if you spend some time looking at the given responses before posting a very similar question. If you do this, I expect you'll also solve this problem yourself and will learn faster, and be happier as a result! –  Derek Jennings Oct 3 '10 at 17:07
I am downvoting this question. You asked for help, and Robin gave it to you. Note that what he advised is exactly the same as what is carried out in muad's answer. Your response to him makes me question whether you asked your question in good faith, hence my downvote. –  Matt E Oct 3 '10 at 18:53
Also, "complex-numbers" is not the correct tag for this question. All the quantities involved are real numbers. –  Matt E Oct 3 '10 at 18:55

You want to evaluate

$$(\sqrt{3 - \sqrt{5}}) (3 + \sqrt{5}) (\sqrt{10} - \sqrt{2})$$

it is equal to

$$\sqrt{3 - \sqrt{5}} \sqrt{(3 + \sqrt{5})^2} \sqrt{(\sqrt{10} - \sqrt{2})^2}$$

merging the top level square roots gives

$$\sqrt{(3 - \sqrt{5}) (3 + \sqrt{5})^2 (\sqrt{10} - \sqrt{2})^2}$$

multiplying this out gives

$$\sqrt{144 - 24 \sqrt{20} + 48 \sqrt{5} - 8 \sqrt{100}}$$

pulling a factor of 48 into the radical to get like terms gives

$$\sqrt{64 - 24 \sqrt{20} + 24 \sqrt{20}}$$

which is $8$.

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Omg, my head is going to explode, finally I understood everything. Thanks. –  hey Oct 3 '10 at 17:10
@muad: I think that the first step requires that the second two factors are positive--they obviously are, but it might be worth noting. –  Isaac Oct 3 '10 at 19:19

Square it, then put $\rm\ \: a = 3,\ b = 2\:a-1 = 5,\ c = 2\ \$ in:

$\rm\quad\quad\quad\quad (a-\sqrt b)\ (a+\sqrt b)^2\ ((\sqrt b - 1)\sqrt c)^2$

$\rm\quad\quad =\ (a^2 \: -\: b)\ \:(a+\sqrt b)\ (b+1-2\sqrt b)\ c$

$\rm\quad\quad =\ (a^2\:-\: b)^2\ 2\:c\ \ \ \:$ via $\rm\:\ \ b+1 \ =\ 2\:a$

NOTE $\:$ Replacing numbers by functions makes it both simpler and more general, as in your prior question

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To the person who downvoted: please explain. If something is not clear then please ask questions and I will elaborate. –  Bill Dubuque Oct 3 '10 at 22:11