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Simplify the following fraction as much as possible $$\def\A{\sqrt{15}}\def\B{\sqrt{35}} \frac{(8+2\A)^{3/2}+(8-2\A)^{3/2}}{(12+2\B)^{3/2}-(12-2\B)^{3/2}} $$

This problem is just driving me insane. I've tried so many different approaches but it keeps becoming a mess for me

I worked with conjugates but that won't help. The first step would be to get rid of the exponent.

Any help on this problem?

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  • $\begingroup$ Have you tried first cubing the Radical Binomials and then square rooting them? $\endgroup$ – thepiercingarrow Oct 6 '15 at 1:39
  • $\begingroup$ do you mean cubing them individually, for example taking $[(8+2\sqrt{15})^{3/2}]^3$ $\endgroup$ – Caddy Heron Oct 6 '15 at 1:54
  • $\begingroup$ No. I mean first cubing $(8+2\sqrt{15})^3$ (spoiler alert - its really big, and you end up with two radical terms). Then you set $(a+b\sqrt{15})^2=(8+2\sqrt{15})^3$ and solve for $a$ and $b$ and do that or all the terms. $\endgroup$ – thepiercingarrow Oct 6 '15 at 11:57
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Hint:

  • What is $\Big(\sqrt5 + \sqrt3\Big)^2$ and $\Big(\sqrt5 - \sqrt3\Big)^2$ ?

  • What is $\Big(\sqrt7 + \sqrt5\Big)^2$ and $\Big(\sqrt7 - \sqrt5\Big)^2$ ?

Another hint:

Please find the following two formulas and use them.

  1. $(a+b)^3 + (a-b)^3$

  2. $(a+b)^3 - (a-b)^3$

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  • $\begingroup$ well for 1. its $5+ 2\sqrt{15}+3=8+2\sqrt{15}$. I see, its the above equation but how did you get that? $\endgroup$ – Caddy Heron Oct 6 '15 at 6:21
  • $\begingroup$ If I tell you the following: The sum of two integers is 8 and the product is 15. Can you (mentally) find the integers? $\endgroup$ – PTDS Oct 6 '15 at 20:20
  • $\begingroup$ Similarly, if I tell you the following: The sum of two integers is 12 and the product is 35. Can you (mentally) find the integers? $\endgroup$ – PTDS Oct 6 '15 at 20:20
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For the nominator I get: $$\sqrt{8+2\sqrt{15}}^3+\sqrt{8-2\sqrt{15}}^3 = 28\sqrt{5}$$ For the denominator I get: $$\sqrt{12+2\sqrt{35}}^3-\sqrt{12-2\sqrt{35}}^3 = 52\sqrt{5}$$ So in the end I get: $$\frac{28\sqrt{5}}{52\sqrt{5}} = \frac{7}{13}$$

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