heureka

A:

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\left[~x \left( x+2 \right) \right~]^3 + 8 x^3 } &\mathbf{
=}& \mathbf{0} \\
x^3 \left( x+2 \right)^3 + 8 x^3 &=& 0\\
x^3 \left( x^3+3x^2\cdot 2 + 3x\cdot 2^2 + 2^3 \right) + 8 x^3 &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 8 \right) + 8 x^3 &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 8 +8\right) &=& 0\\
x^3 \left( x^3+6x^2 + 12x + 16 \right) &=& 0\\
\end{array}
$}}$$

I. First solution

$$x^3=0 \qquad \Rightarrow \qquad x=0$$

II. Second solution

$$\underbrace{x^3+6x^2 + 12x + 16}_{=(x+4)(x^2+2x+4)} = 0 \qquad \Rightarrow \qquad x = -4$$

III. No more solutions

$$x^2+2x+4 = 0$$

The parabola has no solutions, because the vertex of this parabola (-1 | 3) lies above the x-axis

B:

$$\small{\text{$
\begin{array}{rcl}
\mathbf{
\left[~x \left( x+2 \right) \right~]^3 + 8 x^3 } &\mathbf{
=}& \mathbf{0} \\
x^3 \left( x+2 \right)^3 + 8 x^3 &=& 0\\
x^3 \left[~~\left( x+2 \right)^3 + 8~~\right] &=& 0\\
\end{array}
$}}$$

I. First solution:

$$x^3=0 \qquad \Rightarrow \qquad x=0$$

II. Second solution:

$$\small{\text{$
\begin{array}{rcl}
\left( x+2 \right)^3 + 8 &=& 0\\
\left( x+2 \right)^3 &=& -8 \qquad | \qquad \sqrt[3]{}\\
x+2 &=& \sqrt[3]{-8}\\
x+2 &=& -2\\
x &=& -2-2\\
x &=& -4\\
\end{array}
$}}$$

Only one other integer requires nine cubes of positive integers to represent its value as a sum of cubes. This other integer is between 200 and 300. Can you find it ?

$$239=5^3+3^3+3^3+3^3 +2^3+2^3+2^3+2^3+1^3$$

Find the equation of the straight line which is parallel to the line whose equation is 3x + 4y = 0, and which passes through the point of intersection of the lines x - 2y = a and x + 3y = 2a. Express the answer in the form Ax + By = C

I. The slope of the new straight line:

$$\small{\text{$
3x + 4y = 0 \rm{~~or~~}y = -\frac34 \cdot x \rm{~~so~the ~ slope ~is ~~} -\frac34
$.}}
\small{\text{$
\rm{~The ~ slope ~of~a ~parallel ~line ~is ~ also ~} -\frac34
$}}$$

II. intersection of the lines x - 2y = a and x + 3y = 2a:

$$\small{\text{$
\begin{array}{lrcl}
(1): & x - 2y &=& a \\
(2): & x+3y &=& 2a\\
\hline
(2)-(1): & x-x + 3y-(-2y) &=& 2a-a \\
& 3y+2y &=& a \\
& 5y &=& a \\
& \textcolor[rgb]{1,0,0}{y} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac15 a}\\
\hline
(1): & x - 2y &=& a \\
& x - 2y &=& a \\
& x &=& a + 2y \quad | \quad y = \frac15 a \\
& x &=& a + 2(\frac15 a) \\
& \textcolor[rgb]{1,0,0}{x} & \textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac75 a} \\
\end{array}
$}}$$

The intersection is $$(~\frac75 a ~|~ \frac15 a~)$$

III. Find the equation of the straight line:

$$\small{\text{$
\begin{array}{rcl}
y &=&mx+b \qquad | \qquad m = -\frac34 \\
y&=&-\frac34 \cdot x+b
\end{array}
$}}$$

find b:

$$\small{\text{$
\begin{array}{rcl}
y_s&=&-\frac34 \cdot x_s+b \qquad | \qquad x_s = \frac75 a \rm{~~and~~} y_s = \frac15 a\\\\
\frac15 a &=&-\frac34 \cdot \frac75 a+b \\\\
b&=& \frac15 a+\frac34 \cdot \frac75 a \\\\
b&=& \frac{25}{20} a \\\\
\textcolor[rgb]{1,0,0}{b}&\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{\frac{5}{4} a} \\\\
\end{array}
$}}$$

The equation of the straight line is:

$$\small{\text{$
\begin{array}{lrcl}
&y &=& mx+b \qquad | \qquad m = -\frac34 \rm{~~and~~} b = \frac54 a\\\\
&\mathbf{y} &\mathbf{=}& \mathbf{-\frac34 \cdot x + \frac{5}{4} a}\\\\
\rm{or~~} \\\\
&\mathbf{\frac34 \cdot x } + 1\cdot \mathbf{y} &\mathbf{=}& \mathbf{ \frac{5}{4} a }
\end{array}
$}}$$

what is the sum of the interior angles of a polygon with 30 sides ?

How many turquoise squares will be required to build the twentieth figure in this pattern?

n = turquoise squares

n-1 = white squares

n + n - 1  = all squares = $$\small{\text{
$
(2i-1)^2
$}}$$    i is the number of the figure

$$\begin{array}{rcl}
n+n-1 &=& (2i-1)^2\\
2n-1 &=& (2i-1)^2\\
2n &=& 1+(2i-1)^2\\\\
n &=& \dfrac{1+(2i-1)^2}{2}\\
\end{array}\\\\\\
\boxed{~n_i = \dfrac{1+(2i-1)^2}{2} \qquad \rm{or} \qquad$ n_i = 1 + 2i(i-1)$~}\\\\
n_1 = \dfrac{1+(2*1-1)^2}{2} = 1\\\\
n_2 = \dfrac{1+(2*2-1)^2}{2} = 5\\\\
n_3 = \dfrac{1+(2*3-1)^2}{2} = 13\\\\
n_4 = \dfrac{1+(2*4-1)^2}{2} = 25\\\\
\cdots\\\\
n_{20} = \dfrac{1+(2*20-1)^2}{2} = 761 \small{\text{$\qquad \rm{or} \qquad n_{20}= 1+2\cdot 20(20-1) = 1+40\cdot 19=761 $}}$$

Eine Pumpe füllt ein Bassin in drei Stunden, eine andere füllt das Bassin in fünf Stunden. Wie viele Minuten dauert es, bis das Bassin gefüllt ist, wenn beide Pumpen gleichzeitig in Betrieb sind ?

$$\mathbf{
v_{1_{\rm{Pumpe_1}}} = \dfrac{ 1~\rm{Bassin} } { 3~h } \qquad
v_{2_{\rm{Pumpe_2}}} = \dfrac{ 1~\rm{Bassin} } { 5~h }
}\\\\\\
\small{\text{
$
\begin{array}{rcl}
(v_1+v_2)\cdot t &=& 1~\rm{Bassin} \\\\
t &=& \dfrac { 1 } { v_1+v_2 }\\\\
t &=& \dfrac { 1 } { \dfrac{ 1 } { 3~h } + \dfrac{ 1 } { 5~h } }\\\\
t &=& \dfrac { 1 } { \dfrac{ 3~h+5~h } { 15~h^2 } }\\\\
t &=& \dfrac { 15~h^2 } { 8~h }\\\\
t &=& \dfrac { 15~h } { 8 }\\\\
t &=& 1,875~h \\\\
t &=& 1~\rm{Stunde}~ 52,5~\rm{Minuten} \\\\
\end{array}
$}}$$

Beide Pumpen gleichzeitig in Betrieb benötigen 1 Stunde und 52,5 Minuten oder 112,5 Minuten

sieben mal sieben sind neunundvierzig

(m - a)(m - b)(m - c) . . . (m - z) =

how to find a side of the equilateral triangle when only the area A is known!

AE = AB = BE ( equilateral triangle )

P.S.

$$\mathbf{
2A = \overline{AB}\cdot \overline{BE}\cdot \sin{(60\ensurement{^{\circ}})}
}\\\\
\small{\text{
\begin{array}{rcl}
2A &=& \overline{AB}\cdot \overline{BE}\cdot \sin{(60\ensurement{^{\circ}})} \qquad | \qquad \overline{AB} = \overline{BE}\\\\
2A &=& \overline{BE}\cdot \overline{BE}\cdot \sin{(60\ensurement{^{\circ}})}\\\\
2A &=& \overline{BE}^2\cdot \sin{(60\ensurement{^{\circ}})}
\qquad | \qquad A = \sin{ (60\ensurement{^{\circ}}) } \\\\
2 \cdot \sin{ (60\ensurement{^{\circ}}) } &=& \overline{BE}^2\cdot \sin{(60\ensurement{^{\circ}})} \\\\
2 &=& \overline{BE}^2 \\\\
\overline{BE}^2 &=& 2 \qquad | \qquad \sqrt{} \\\\
\overline{BE} &=& \sqrt{2}
\end{array}
$
$}}$$