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First, note that $m^3+n^3 = (m+n)(m^2+n^2-mn)$ and $(4+\sqrt {15})(4-\sqrt {15})=(6+\sqrt {35})(6-\sqrt {35})=1$

Let x = $\frac {(4+\sqrt {15})^{\frac {3}{2}}+ (4-\sqrt {15})^{\frac {3}{2}}}{(6+\sqrt {35})^{\frac {3}{2}}-(6-\sqrt {15})^{\frac {3}{2}}}$

Squaring, $x^2=$ $\frac {(4+\sqrt {15})^3+ (4-\sqrt {15})^3+2[(4+\sqrt {15})(4-\sqrt {15})]^{\frac {3}{2}}}{(6+\sqrt {35})^3-(6-\sqrt {15})^3-2[(6+\sqrt {15})(6-\sqrt {15})]^{\frac {3}{2}}}$ $=\frac {8 (16+8\sqrt {15}+15+16-8\sqrt {15}+15-1)+2}{12 (36+12\sqrt {35}+35+36-12\sqrt {35 }+35-1)-2}$ $=\frac {8*61+2}{12*141-2}$ $=\frac {49}{169}$ $\Rightarrow x=\frac {7}{13}$ since x is positive. $a+b=7+13=\boxed {20}$

We note that: $\left(\dfrac{3+\sqrt{15}}{\sqrt{6}}\right)^2 = \dfrac{24 + 6\sqrt{15}}{6} = 4+\sqrt{15}$

\(\begin{array} {} \text{Similarly,} & 4-\sqrt{15} = \left(\dfrac{\sqrt{15}-3}{\sqrt{6}}\right)^2, & 6+\sqrt{35} = \left(\dfrac{\sqrt{35}+5}{\sqrt{10}}\right)^2, & 6-\sqrt{35} = \left(\dfrac{\sqrt{35}-5}{\sqrt{10}} \right)^2 \end{array} \)

Therefore,

$\begin{aligned} \frac{(4+\sqrt{15})^{\frac{3}{2}} + (4-\sqrt{15})^{\frac{3}{2}}}{(6+\sqrt{35})^{\frac{3}{2}} - (6-\sqrt{35})^{\frac{3}{2}}} & = \frac{(\sqrt{15}+3)^3 + (\sqrt{15}-3)^3}{(\sqrt{35}+5)^3 - (\sqrt{35}-5)^3}\times \frac{(\sqrt{10})^3}{(\sqrt{6})^3} \\ & = \frac{\color{#3D99F6}{2(15\sqrt{15}+27\sqrt{15})}}{\color{#D61F06}{2(525+125)}}\times \frac{10\sqrt{10}}{6\sqrt{6}} \quad \quad \small \color{#3D99F6}{\text{Even}} \text{ or } \color{#D61F06}{\text{odd}} \text{ terms cancel out.} \\ & = \frac{42\sqrt{15}}{650} \times \frac{10\sqrt{5}}{6\sqrt{3}} = \frac{7}{13} \end{aligned}$

$\Rightarrow a + b = 7 + 13 = \boxed{20}$

An easy method is to realize $4±√15=(√3/2 ± √5/2)^2 and 6±√35=(√7/2 ± √5/2)^2$ then numerator simplifies to √10 (7) while denominator simplifies to √10 (13) , answer is therefore 7+13=20

First Note that,
$(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}})=(\sqrt{6+\sqrt{35}}-\sqrt{6-\sqrt{35}})=\sqrt{10}$
Proving.

Let $(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}})=x$ and $(\sqrt{6+\sqrt{35}}-\sqrt{6-\sqrt{35}})=y$

Squring both sides.

$x^2=8+(2×1)$ and $y^2=12-(2×1)$
$x=\sqrt{10}$ and $y=\sqrt{10}$
$\Rightarrow \boxed{x=y}.$

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$\Rightarrow \dfrac{(\sqrt{4+\sqrt{15}})^3+(\sqrt{4-\sqrt{15}})^3}{(\sqrt{6+\sqrt{35}})^3-(\sqrt{6-\sqrt{35}})^3}$

Using $(a^3+b^3)$ and $(a^3-b^3)$ identities.

$=\dfrac{(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}})(4+\sqrt{15}+4-\sqrt{15}-\sqrt{16-15}}{(\sqrt{6+\sqrt{35}}-\sqrt{6-\sqrt{35}})(6+\sqrt{35}+6-\sqrt{35}+\sqrt{36-35}}$

$=\dfrac{(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}})×(7)}{(\sqrt{6+\sqrt{35}}-\sqrt{6-\sqrt{35}})×(13)}=\dfrac{a}{b}$

$=\dfrac{\sqrt{10}×7}{\sqrt{10}×13}=\dfrac{a}{b}$

$\Rightarrow \boxed{\dfrac{7}{13}=\dfrac{a}{b}}$

$\Rightarrow a+b=7+13=\boxed{20}.$