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How do we go from a general form to a canonique form (2nd degree)

see: https://www.youtube.com/watch?v=5Yke-3yNYcA

Suppose  and .

Then, what is the average (arithmetic mean) of the three products  , , and    ?

$$\small{\text{
$
\begin{array}{rcl}
(p+q+r)^2&=&(p+q+r)(p+q+r)=p^2+q^2+r^2+2(pq+qr+rp)\\
(p+q+r)^2&=&p^2+q^2+r^2+2(pq+qr+rp)\\
7^2&=&9+2(pq+qr+rp)\\
2(pq+qr+rp)&=&7^2-9=49-9=40\\
(pq+qr+rp) &=& 20
\end{array}
$
}}$$

the average (arithmetic mean) of the three products  , , and    ?   $$\frac{(pq+qr+rp) }{3} = \frac{20}{3}=6\frac{2}{3}$$

.

see: http://www.intmath.com/trigonometric-functions/2-sin-cos-tan-csc-sec-cot.php

The sequence , , , . . ., has the property that  for all .

If , then determine 

$$\small{\text{
$
\begin{array}{rcrcrcr}
x_{11}&=& x_{10}+x_{9} &=& (x_9+x_8)+x_9 &=& 2x_9+x_8\\
&=& 2x_9+x_8 &=& 2(x_8+x_7)+x_8 &=& 3x_8+2x_7\\
&=& 3x_8+2x_7 &=& 3(x_7+x_6)+2x_7 &=& 5x_7+3x_6\\
&=& 5x_7+3x_6 &=& 5(x_6+x_5)+3x_6 &=& 8x_6+5x_5\\
&=& 8x_6+5x_5 &=& 8(x_5+x_4)+5x_5 &=& 13x_5+8x_4\\
&=& 13x_5+8x_4 &=& 13(x_4+x_3)+8x_4 &=& 21x_4+13x_3\\
&=& 21x_4+13x_3 &=& 21(x_3+x_2)+13x_3 &=& 34x_3+21x_2\\
&=& 34x_3+21x_2 &=& 34(x_2+x_1)+21x_2 &=& 55x_2+34x_1\\
\end{array}
$
}}\\
\small{\text{$x_{11} = 55x_2+34x_1$}}$$

 

$$\small{\text{$
\begin{array}{rcl}
55x_2+34x_1 -x_1 &=& 99 \\
55x_2 +33x_1&=&99 \quad | \quad :11 \\
\boxed{5x_2+3x_1 = 9}
\end{array}
$
}}$$

$$\small{\text{
$
\begin{array}{rcrcr}
x_3 &=& &=& x_2+x_1 \\
x_4 &=& x_3+x_2 &=& 2x_2+x_1 \\
x_5 &=& x_4+x_3 &=& 3x_2+2x_1 \\
x_6 &=& x_5+x_4 &=& 5x_2+3x_1 \\
\end{array}
$
}}\\
\small{\text{$x_6= 5x_2+3x_1 $}}=9$$

.

Evaluate

  

if

 .

$$\left(a+\frac{1}{a}\right)^3=6^3\\\\
a^3+3a^2*\frac{1}{a}+3a*\frac{1}{a^2}+\frac{1}{a^3}=6^3\\\\
a^3+3a+3\frac{1}{a}+\frac{1}{a^3}=6^3\\\\
a^3+\frac{1}{a^3}+3a+3\frac{1}{a}=6^3\\\\
a^3+\frac{1}{a^3}+3\underbrace{\left(a+\frac{1}{a}\right)}_{=6}=6^3\\\\
a^3+\frac{1}{a^3}+3*6=6^3\\\\
a^3+\frac{1}{a^3}=6^3-3*6\\\\
a^3+\frac{1}{a^3}=6(6^2-3)\\\\
a^3+\frac{1}{a^3}=6(33)\\\\
a^3+\frac{1}{a^3}=198$$

.