For the real numbers $a$ and $b$ satisfying
$$a ^ { 3 } = \frac { 1 } { \sqrt { 5 } - 2 } , \quad b ^ { 3 } = 2 - \sqrt { 5 }$$
we are to find the value of $a + b$.
When we set $x = a + b$, we have
$$x ^ { 3 } = ( a + b ) ^ { 3 } = a ^ { 3 } + b ^ { 3 } + \mathbf { A } a b ( a + b ) .$$
Since $a b = \mathbf { B C }$, we know that this $x$ satisfies
$$x ^ { 3 } + \mathbf { D } x - \mathbf { E } = 0 .$$
The left side of this equation can be factorized as follows:
$$\begin{aligned} x ^ { 3 } + \mathbf { D } x - \mathbf { E } & = \left( x ^ { 3 } - \mathbf { F } \right) + \mathbf { D } \left( x - \frac { \mathbf { F } } { \mathbf { F } } \right) \\ & = ( x - \mathbf { F } ) \left( x ^ { 2 } + x + \mathbf { G } \right) . \end{aligned}$$
Since
$$x ^ { 2 } + x + \mathbf { G } = \left( x + \frac { \mathbf { H } } { \mathbf { I } } \right) ^ { 2 } + \frac { \mathbf { J K } } { \mathbf { L } } > 0 ,$$
we obtain $x = a + b = \mathbf { M }$.

Consider the two functions $y = x ^ { 2 } + a x + a$ and $y = x + 1$.
(1) The number of points at which the graphs of the two functions meet depends on the relationship of $a$ with the numbers $\mathbf { Q }$ and $\mathbf { R }$ in the following way: (For $\mathbf { N } \sim \mathbf { P }$ choose which of (0) $\sim$ (2) gives the correct condition for the question.)
(i) The condition under which the graphs of the two functions intersect at two different points is $\mathbf { N }$.
(ii) The condition under which the graphs of the two functions are tangent at a point is $\mathbf{O}$.
(iii) The condition under which the graph of $y = x ^ { 2 } + a x + a$ is always above the graph of $y = x + 1$ is $\mathbf { P }$.
$$\begin{aligned} & \text { (0) } \mathrm { Q } < a < \mathrm { R } \\ & \text { (1) } a = \mathrm { Q } \text { or } a = \mathrm { R } \\ & \text { (2) } a < \mathrm { Q } \text { or } \mathrm { R } < a \end{aligned}$$
(2) Let us consider the case where the value of $a$ satisfies P. Let $g ( x )$ be the difference between the values of the two functions, so $g ( x ) = x ^ { 2 } + a x + a - ( x + 1 )$, and let $m$ be the minimum value of $g ( x )$. Then
$$m = - \frac { \mathbf { S } } { \mathbf { T } } \left( a ^ { 2 } - \mathbf { U } a + \mathbf { U } \right)$$
Hence $m$ takes the maximum at $a = \mathbf { W }$ and its value there is $m = \mathbf { W }$.

There are six boxes numbered from 1 to 6. We are to put four balls of different sizes into these boxes.
(1) There are altogether $\mathbf{AA}$ ways to put the four balls into the boxes.
(2) There are $\mathbf{CDE}$ ways to put the four balls into four separate boxes.
(3) There are $\mathbf{FGH}$ ways to put three balls into one box and the fourth ball into another.
(4) There are $\mathbf{IJK}$ ways to put at least one ball into the box numbered 1.

Consider the quadratic function of $x$
$$x ^ { 2 } + ( 4 a - 6 ) x + 2 a + b + 5 = 0 . \tag{1}$$
We are to find the conditions under which one solution is $-1$ and the other solution satisfies the inequality
$$| x + 2 a | < a + 1 . \tag{2}$$
(1) The condition for equation (1) to have the solution $-1$ is
$$b = \mathbf { L } a - \mathbf { M N } .$$
Denote the other solution by $\alpha$. When $\alpha$ is expressed in terms of $a$, we have
$$\alpha = \mathbf { O P } a + \mathbf { Q } .$$
(2) When $a > \mathbf{RS}$, the inequality (2) has a solution, and its solution is
$$\mathbf { TU } a - \mathbf { VV } < x < - a + \mathbf { W } .$$
Hence the conditions are: that $a$ and $b$ satisfy (3) and that $a$ satisfies
$$\mathbf { X } < a < \mathbf { Y } .$$

Consider two quadratic functions
$$\begin{aligned} & y = 2 x ^ { 2 } + 3 a x + 4 b \tag{1}\\ & y = b x ^ { 2 } + c x + d \tag{2} \end{aligned}$$
whose graphs are mutually symmetric with respect to the origin.
(1) From the symmetry with respect to the origin we see that
$$b = \mathbf { AB } , \quad c = \mathbf { C } a , \quad d = \mathbf { D } .$$
Hence (2) can be reduced to
$$y = \mathbf{AB} x ^ { 2 } + \mathbf{C} a x + \mathbf{D} . \tag{3}$$
(2) Let $0 < a < 1$, and consider the graph of (3).
When the range of values of $x$ is $0 \leqq x \leqq \frac { 3 } { 2 }$, the range of values of $y$ in (3) is
$$\frac { \mathbf { E } } { \mathbf { F } } a ^ { 2 } - \frac { \mathbf { G } } { \mathbf { H } } \leq y \leqq \frac { \mathbf { I } } { \mathbf{J}} a ^ { 2 } + \mathbf { K }$$
(3) For any value of $a$, the vertex of the graph of (3) is on the graph of the quadratic function
$$y = \mathbf { L } x ^ { 2 } + \mathbf { M } .$$

Consider a triangle ABC where
$$\mathrm { AB } = 10 , \quad \angle \mathrm { B } = 30 ^ { \circ }$$
and the radius of its inscribed circle O is 1.
(1) Set $a = \mathrm { BC }$ and $b = \mathrm { CA }$. Finding the area $S$ of the triangle ABC by two different methods, we have
$$S = \frac { \mathbf { A } } { \mathbf{B} } \mathbf { B } ,$$
and
$$S = \frac { \mathbf { C } } { \mathbf{C} } \mathbf { D } ( a + b + \mathbf { E C } ) .$$
Hence we obtain
$$b = \mathbf { G } a - \mathbf { H I } .$$
Since the relationship between $a$ and $b$ can also be expressed by the equation
$$b ^ { 2 } = a ^ { 2 } - \mathbf { J K } \sqrt { \mathbf { L } } a + \mathbf { M N O } ,$$
we have
$$a = \frac { \mathbf { P Q } - \mathbf { R } } { 3 } \sqrt { \mathbf { S } } , \quad b = \frac { \mathbf { T U } } { \mathbf{P} } \cdot \frac { \mathbf { V } } { 3 } \sqrt { \mathbf { W } } .$$
(2) Let D denote the point of intersection of the segment BC and the straight line which passes through the two points A and O. We denote the area of the triangle OBC by $S ^ { \prime }$. Since
$$S : S ^ { \prime } = \mathbf { X } : 1 ,$$
it follows that
$$\mathrm { AO } : \mathrm { OD } = \mathbf { Y } : 1 .$$

For the real numbers $a$ and $b$ satisfying
$$a ^ { 3 } = \frac { 1 } { \sqrt { 5 } - 2 } , \quad b ^ { 3 } = 2 - \sqrt { 5 }$$
we are to find the value of $a + b$.
When we set $x = a + b$, we have
$$x ^ { 3 } = ( a + b ) ^ { 3 } = a ^ { 3 } + b ^ { 3 } + \mathbf { A } a b ( a + b ) .$$
Since $a b = \mathbf { B C }$, we know that this $x$ satisfies
$$x ^ { 3 } + \mathbf { D } x - \mathbf { E } = 0 .$$
The left side of this equation can be factorized as follows:
$$\begin{aligned} x ^ { 3 } + \mathbf { D } x - \mathbf { E } & = \left( x ^ { 3 } - \mathbf { F } \right) + \mathbf { D } \left( x - \frac { \mathbf { F } } { \mathbf { E } } \right) \\ & = ( x - \mathbf { F } ) \left( x ^ { 2 } + x + \mathbf { G } \right) . \end{aligned}$$
Since
$$x ^ { 2 } + x + \frac { \mathbf { G } } { \mathbf { H } } = \left( x + \frac { \mathbf { H } } { \mathbf { I } } \right) ^ { 2 } + \frac { \mathbf { J K } } { \mathbf { L } } > 0 ,$$
we obtain $x = a + b = \mathbf { M }$.

Consider the two functions $y = x ^ { 2 } + a x + a$ and $y = x + 1$.
(1) The number of points at which the graphs of the two functions meet depends on the relationship of $a$ with the numbers $\mathbf { Q }$ and $\mathbf { R }$ in the following way: (For $\mathbf { N } \sim \mathbf { P }$ choose which of (0) $\sim$ (2) gives the correct condition for the question.)
(i) The condition under which the graphs of the two functions intersect at two different points is $\mathbf { N }$.
(ii) The condition under which the graphs of the two functions are tangent at a point is $\mathbf{O}$.
(iii) The condition under which the graph of $y = x ^ { 2 } + a x + a$ is always above the graph of $y = x + 1$ is $\mathbf { P }$.
$$\begin{aligned} & \text { (0) } \mathrm { Q } < a < \mathrm { R } \\ & \text { (1) } a = \mathrm { Q } \text { or } a = \mathrm { R } \\ & \text { (2) } a < \mathrm { Q } \text { or } \mathrm { R } < a \end{aligned}$$
(2) Let us consider the case where the value of $a$ satisfies P. Let $g ( x )$ be the difference between the values of the two functions, so $g ( x ) = x ^ { 2 } + a x + a - ( x + 1 )$, and let $m$ be the minimum value of $g ( x )$. Then
$$m = - \frac { \mathbf { S } } { \mathbf { T } } \left( a ^ { 2 } - \mathbf { U } a + \mathbf { U } \right)$$
Hence $m$ takes the maximum at $a = \mathbf { W }$ and its value there is $m = \mathbf { W }$.

Take four points
$$\mathrm { A } ( 1,0 ) , \quad \mathrm { B } ( 0,1 ) , \quad \mathrm { C } ( 3,0 ) , \quad \mathrm { D } ( 0,2 )$$
on a plane with the coordinate system having the origin O. Take two points P and Q on segments AB and CD respectively, such that
$$\mathrm { AP } : \mathrm { PB } = \mathrm { CQ } : \mathrm { QD } = k : 2 .$$
We are to find the minimum possible length of the segment PQ.
(1) First, let us find the value of $x + 2 y$, where $\overrightarrow { \mathrm { PQ } } = ( x , y )$.
Since
$$\overrightarrow { \mathrm { OP } } = \frac { \square \mathbf { A } \overrightarrow { \mathrm { OA } } + k \overrightarrow { \mathrm { OB } } } { k + \mathbf { B } } , \quad \overrightarrow { \mathrm { OQ } } = \frac { \square \mathbf { C } \overrightarrow { \mathrm { OC } } + k \overrightarrow { \mathrm { OD } } } { k + \mathbf { D } } ,$$
we have
$$( x , y ) = \frac { 1 } { k + \mathbf { E } } ( \mathbf { F } , k ) ,$$
which gives $x + 2 y = \mathbf { G }$.
(2) When we represent $\mathrm { PQ } ^ { 2 }$ in terms of $y$, we have
$$\mathrm { PQ } ^ { 2 } = \mathbf { H } y ^ { 2 } - \mathbf { I } y + \mathbf { J } .$$
Hence PQ takes the minimum at $y = \frac { \mathbf { K } } { \mathbf { L } }$ and its value there is $\mathrm { PQ } = \frac { \square \mathbf { M } \sqrt { \mathbf { N } } } { \mathbf { O } }$. In this case, the value of $k$ is $k = \square \mathbf { P }$.

We have a triangle ABC such that
$$\mathrm { AB } = 9 , \quad \mathrm { BC } = 12 , \quad \angle \mathrm { ABC } = 90 ^ { \circ } .$$
There are also two circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ with radii of length $2r$ and $r$, respectively. The two circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ are tangential to each other. Further, $\mathrm { O } _ { 1 }$ is tangent to the two sides AB and AC, and $\mathrm { O } _ { 2 }$ is tangent to the two sides CA and CB. We are to find the value of $r$.
First, let D and E denote the points at which the segment AC is tangent to the circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ respectively, and set $\alpha = \angle \mathrm { O } _ { 1 } \mathrm { AC }$. Then, since $\tan 2 \alpha = \frac { \square \mathbf { A } } { \square }$, we have $\tan \alpha = \frac { \mathbf { C } } { \mathbf { D } }$ using the double-angle formula. Thus, we obtain $\mathrm { AD } = \mathbf { E }$.
Next, set $\beta = \angle \mathrm { O } _ { 2 } \mathrm { CA }$. Since $\alpha + \beta = \mathbf { F G } ^ { \circ }$, we have $\tan \beta = \frac { \mathbf { H } } { \square \mathbf{I} }$ using the addition theorem. Thus, we obtain $\mathrm { CE } = \square r$.
Moreover, it follows that $\mathrm { AC } = \mathbf { K L }$ and $\mathrm { DE } = \mathbf { M } \sqrt { \mathbf { N } } r$.
Finally we obtain
$$r = \frac { \mathbf { O P } ( \mathbf { Q } - \mathbf { R } \sqrt { \mathbf { S } } ) } { 41 }$$

Let $f ( x ) = 4 \sqrt { 3 } e ^ { - x } \cos x + 6 e ^ { - x }$.
(1) Let $a$ and $b$ ($a < b$) be the values of $x$ satisfying $f ( x ) = 0$ on $0 \leqq x < 2 \pi$. Then,
$$a = \frac { \mathbf{A} } { \mathbf{B} } \pi , \quad b = \frac { \mathbf{C} } { \mathbf{D} } \pi$$
(2) The values of the constants $p$ and $q$ satisfying
$$\frac { d } { d x } \left( p e ^ { - x } \cos x + q e ^ { - x } \sin x \right) = e ^ { - x } \cos x$$
are given by
$$p = \frac { \mathbf { E F } } { \mathbf { G } } , \quad q = \frac { \mathbf { H } } { \mathbf { I } } .$$
(3) Using the values of $a$ and $b$ obtained in (1), we set $A = e ^ { - a }$ and $B = e ^ { - b }$. When we calculate the value of $\int _ { a } ^ { b } f ( x ) d x$, we obtain
$$\int _ { a } ^ { b } f ( x ) d x = ( \mathbf { J } - \sqrt { \mathbf{J} } \mathbf { K } ) A - ( \mathbf { L } + \sqrt { \mathbf{L} } ) B .$$

Consider the definite integral $S = \int _ { 0 } ^ { a } x \sqrt { \frac { 1 } { 3 } x + 2 } \, d x$.
(1) Set $t = \sqrt { \frac { 1 } { 3 } x + 2 }$. Then we have
$$\begin{aligned} \int x \sqrt { \frac { 1 } { 3 } x + 2 } \, d x & = \mathbf { N O } \int \left( t ^ { \mathbf { P } } - \mathbf { Q } t ^ { \mathbf { R } } \right) d t \\ & = \mathbf { S } + C , \end{aligned}$$
where $C$ is the integral constant.
(2) Using the result in (1), we have
$$S = \mathbf { T } .$$
Thus we obtain
$$\lim _ { a \rightarrow \infty } \frac { S } { a ^ { \frac { \mathbf { U } } { \mathbf{V} } } } = \frac { \mathbf { W } \sqrt { \mathbf { X } } } { \mathbf { Y Z } }$$
For $\mathbf{S}$ and $\mathbf{T}$, choose the appropriate expression from among the choices (0) $\sim$ (9) below.
(0) $\frac { 6 } { 5 } t ^ { 5 } \left( 3 t ^ { 2 } - 10 \right)$
(1) $\frac { 6 } { 5 } t ^ { 3 } \left( 3 t ^ { 2 } - 10 \right)$
(2) $\frac { 12 } { 5 } t ^ { 5 } \left( 3 t ^ { 2 } - 5 \right)$
(3) $\frac { 12 } { 5 } t ^ { 3 } \left( 3 t ^ { 2 } - 5 \right)$
(4) $\frac { 6 } { 5 } t ^ { 3 } \left( 3 t ^ { 2 } - 5 \right)$
(5) $\frac { 6 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 5 } ( a - 4 ) + 8 \sqrt { 2 } \right\}$
(6) $\frac { 12 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 3 } ( a - 2 ) + 4 \sqrt { 2 } \right\}$
(7) $\frac { 12 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 5 } ( a - 2 ) + 4 \sqrt { 2 } \right\}$
(8) $\frac { 6 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 3 } ( a - 4 ) + 8 \sqrt { 2 } \right\}$
(9) $\frac { 6 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 3 } ( a - 2 ) + 8 \sqrt { 2 } \right\}$