MaxWong

$ab^4 = 384\\ a^2 b^5 = 4608$

Considering the second equation: 

$ab^4 \cdot ab = 4608\\ ab = \dfrac{4608}{384} = 12$

Considering the first equation:

$ab^4 = 384\\ ab\cdot b^3 = 384\\ b = \sqrt[3]{\dfrac{384}{12}} = 2\sqrt[3]4$

Substituting directly gives $a =3 \sqrt[3]{2}$

$\sqrt{8} = \sqrt{2^2 \cdot 2} = 2 \sqrt2\\ \sqrt{32} = \sqrt{4^2 \cdot 2} = 4\sqrt 2$

So

$\quad\sqrt8 + 3\sqrt 2 + \sqrt{32}\\ = 2\sqrt 2 + 3\sqrt 2 + 4\sqrt 2\\ = 9\sqrt 2$

By considering power of points at X,

$2(2 + 7) = 3(3 + ED)\\ ED = 3$

$\quad\left(\sqrt{3x} + \sqrt 5\right)\left(\sqrt{15x} + 2\sqrt{30}\right)\\ = \sqrt{45x^2} + \sqrt{75x} + 2\sqrt{90x} + 2\sqrt{150}\\ =3x\sqrt5 + 5\sqrt{3x} + 6\sqrt{10x} + 10\sqrt 6$

$f(f^{-1}(x)) = f(4)\\ x = 3\cdot 4^3 + 2 = 194$

$\dfrac15 \div 4 = \dfrac15 \div \dfrac41 = \dfrac15 \times \dfrac14 = \dfrac{1\times 1}{5\times 4} = \dfrac1{20}$

Everything outside on the surface of the cube is visible.

That means the taking away the outside layer, the inner 8 x 8 x 8 cube would not be visible.

Total number of visible 1 x 1 x 1 cubes = $10\times 10 \times 10 - 8 \times 8 \times 8 = \boxed{488}$

First, we find $f^{-1}(x)$ in terms of a and b.

$f(x) = ax + b\\ x = af^{-1}(x) + b\\ f^{-1}(x) = \dfrac{x - b}a$

Now,

$g(x) = f^{-1}(x) - 3\\ 5x - 4 = \dfrac xa - \left(\dfrac ba + 3\right)$

Comparing coefficients:

$\dfrac1a = 5\\ a = \dfrac15$

$5b + 3 = 4\\ b = \dfrac15$

Therefore

$5a + 5b = 5\cdot \dfrac15 + 5\cdot \dfrac15 = 2$

Adding the 5 equations:

4(a + b + c + d + e) = 84

a + b + c + d + e = 21 --- (*)

Subtracting each equation from (*),

e = 21 - 14 = 7

d = 21 - 15 = 6

c = 21 - 17 = 4

b = 21 - 18 = 3

a = 21 - 20 = 1

abcde = 1(3)(4)(6)(7) = 504. 

Using the definition:

$\cos B = \dfrac6{10} = \dfrac{AB}{BC}$

From this, we can conclude AB = 6.

By Pythagorean theorem, AC = 8.

$\tan C = \dfrac{AB}{AC} = \dfrac68 = \boxed{\dfrac34}$