Roots!

Given that,

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

After squaring both sides and we get,

$(4+\sqrt{15})^3+(4-\sqrt{15})^3-2((4+\sqrt{15})(4-\sqrt{15}))^\frac{3}{2}=6\times(K^2)$

now,we know

$\boxed{(a+b)(a-b)=a^2-b^2}$

$\boxed{(a+b)^3=a^3+3a^2b+3ab^2+b^3}$

$\boxed{(a-b)^3=a^3-3a^2b+3ab^2-b^3}$

$\boxed{(a+b)^3+(a-b)^3=2a^3+6ab^2}$

then,applying these formulas in the equation we get,

$(2\times4^3)+(6\times4\times(\sqrt{15})^2)-(2\times(4^2-(\sqrt{15})^2)^\frac{3}{2})=6\times(K^2)$

$(2\times64)+(6\times4\times15)-(2\times1^\frac{3}{2})=6\times(K^2)$

$128+360-2=6\times(K^2)$

$6\times(K^2)=486$

$K^2=486/6$

$K^2=81$

$K=\sqrt{81}$

$K=\boxed{9}$

so, 9 is the answer

Let, (4+√15)^(1/2) be a, Then,(4-√15)^(1/2) will be 1/a = b (say) {Just multiply numerator and denominator by 4+√15} So, a b = 1. Now the given problem becomes : a^3 - b^3 Use the identity : a^3 - b^3 = (a-b) (a^2+b^2+a b) Then, a^2+b^2+a b becomes 4+√15+4-√15+1 = 9. And a-b becomes (4+√15)^1/2 - (4-√15)^1/2 Square it and then take square root of this expression Which basically means : (a^2)^1/2 = a(This is done to easily solve it) i.e. (a-b)^2 = 4+√15+4-√15-2 = 6. Now take square root i.e.
√(a-b)^2 = √6 Thus, it becomes 9 √6 Comparing with K √6, we get K=9.

Same concept with Sagar Maity . Maybe better presented.

Let $a = 4+\sqrt{15}$ and $b = 4-\sqrt{15}$ .

Then we note that $a+b=8$ and $ab= 4^2-(\sqrt{15})^2=16-15=1$ .

Then, $\space LHS=a^{\frac{3}{2}} - b^{\frac{3}{2}}$

$\Rightarrow LHS^2 = a^3-2(ab)^{\frac{3}{2}} + b^3 = a^3 + b^3 - 2(1)$

$\quad \quad \quad \quad = (a+b)(a^2-ab+b^2)-2 = 8(a^2+2ab+b^2-3ab)-2$

$\quad \quad \quad \quad = 8[(a+b)^2-3]-2 = 8(64-3)-2 = 486$

$\Rightarrow (K\sqrt{6})^2 = 6K^2 = 486\quad \Rightarrow K^2=81\quad \Rightarrow K = \boxed{9}$