MaxWong

Divide by 5 on both sides.

$125x = \dfrac 95$

Note that $(\sin (a))^5 < 1$ and $(\cos(a))^5 < 1$.

Then, we have

$\begin{array}{rcl} (\sin(a))^7 + (\cos(a))^7 &=& (\sin(a))^2(\sin(a))^5 + (\cos(a))^2 (\cos(a))^5\\ &<& (\sin(a))^2 + (\cos(a))^2\\ &=& 1 \end{array}$

The inequality is shown.

Suppose the radius of the circle = r units 

$\pi r^2 = 25\pi\\ r = 5$

Then, the radius of the sphere is $\sqrt{r^2 + 5^2} = 5 \sqrt 2$ units.

Then the surface area is $4 \pi (5 \sqrt 2)^2 = 200\pi$.

Using Pythagoras theorem in $\triangle ABX$ gives $AX^2 + BX^2 = AB^2$, substituting AX = 15 and AB = 25 gives

$15^2 + BX^2 = 25^2\\ BX^2 = 400\\ BX = 20$

Note that $\angle BAD = \angle CAD = \dfrac12 \cdot 30^\circ = 15^\circ$.

Considering the interior angle sum of $\triangle ADC$ gives $\angle ADC = 120^\circ$.

Using Sine Law in $\triangle ADC$ gives 

$\dfrac{10}{\sin 120^\circ} = \dfrac{AD}{\sin 45^\circ}\\ AD = \dfrac{10 \sqrt 6}3$

Considering the angles around point D gives $\angle ADB = 60^\circ$. Considering the interior angle sum of $\triangle ABD$ gives $\angle ABD = 105^\circ$.

Using Sine Law in $\triangle ABC$ gives 

$\dfrac{10}{\sin 105^\circ} = \dfrac{AB}{\sin 45^\circ}\\ AB = 10(\sqrt 3 - 1)$

Then, the area of $\triangle ABD$ is $\dfrac12 (AB)(AD) \sin 15^\circ = \dfrac12 (10(\sqrt 3 - 1))\cdot \dfrac{10 \sqrt 6}3 \cdot \dfrac{\sqrt 6 - \sqrt 2}4 = \dfrac{50(2 \sqrt 3 - 3)}{3}$.

Each of the 8 pyramids being cut off has a right isosceles base with equal sides 3, and height 3.

Using the formula for volume of cube and volume of pyramid, we have

$\text{Volume} = 6^3 - 8 \cdot \dfrac13 \left(\dfrac{3 \cdot 3}2\right)\cdot 3 = 180$

$\displaystyle \frac{1}{\sqrt{36} + \sqrt{42}} + \frac{1}{\sqrt{39} + \sqrt{45}}\\ =\dfrac{\sqrt{42} - \sqrt{36}}{42 - 36} + \dfrac{\sqrt{45} - \sqrt{39}}{45 - 39}\\ = \dfrac16 \left(\sqrt{45} + \sqrt{42} - \sqrt{39} - \sqrt{36}\right)\\ = \dfrac{3\sqrt 5 + \sqrt{42} - \sqrt{39} - 6}6$

$\dfrac1{1 + \dfrac1{4 + \dfrac12}}\\ = \dfrac1{1 + \dfrac1{\dfrac92}}\\ = \dfrac1{1 + \dfrac29}\\ = \dfrac1{\dfrac{11}{9}}\\ = \dfrac9{11}$

Recall that 

If the roots of the quadratic equation $x^2 + bx + c = 0$ are $\alpha$ and $\beta$, then $\alpha + \beta = -b$ and $\alpha \beta = c$.

(Vieta's formula)

By this fact, we have $b = -(5 + 3i) - (1 - i) = -6 - 2i$, $c = (5 + 3i)(1 - i) = 5 - 2i - 3i^2 = 8 - 2i$.

Then, $b + c = -6-2i + (8-2i) = \boxed{2 - 4i}$

Suppose $xy = a$. Then,

$x^2 + y^2 = (x^2 + 2xy + y^2) - 2xy = (x + y)^2 - 2xy = 100 - 2a$

$x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1000 - 30a$

So the second equation is equivalent to 

$100 - 2a = 1000 - 30a\\ 28a = 900\\ a = \dfrac{225}7\\ xy = \dfrac{225}7\\ x = \dfrac{225}{7y}$

Substituting this into x + y = 10, we have

$y + \dfrac{225}{7y} = 10\\ 7y^2 - 70y + 225 = 0$

Checking the discriminant of this equation, $\Delta = (-70)^2 - 4(7)(225) = -1400 < 0$.

Therefore, there are no real solutions.