# Quod Erat Demonstrandum

## 2015/09/23

### 兩數相等

Filed under: Fun — johnmayhk @ 5:43 下午
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$8\times 9=9\times 8$

$\Rightarrow 8\times (17-8)=9\times (17-9)$

$\Rightarrow 8\times 17-8^2=9\times 17-9^2$

$\Rightarrow 8^2-8\times 17=9^2-9\times 17$

$\Rightarrow 8^2-8\times 17+(\frac{17}{2})^2=9^2-9\times 17+(\frac{17}{2})^2$

（配方法了）

$\Rightarrow (8-\frac{17}{2})^2=(9-\frac{17}{2})^2$

$\Rightarrow 8-\frac{17}{2}=9-\frac{17}{2}$

$\Rightarrow 8=9$

$a=x-y$
$b=(x+y)^2$
$c=4(x^2-xy+y^2)$

$(a+\sqrt{b-c})^3$

$=a^3+3a^2\sqrt{b-c}+3a(b-c)+(b-c)\sqrt{b-c}$

$=a^3+3a(b-c)+(3a^2+b-c)\sqrt{b-c}$

$3a^2+b-c$

$=3(x-y)^2+(x+y)^2-4(x^2-xy+y^2)$

$=3x^2-6xy+3y^2+x^2+2xy+y^2-4x^2+4xy-4y^2$

$=0$

$(a+\sqrt{b-c})^3=a^3+3a(b-c)$　………………. (1)

$(a-\sqrt{b-c})^3$

$=a^3-3a^2\sqrt{b-c}+3a(b-c)-(b-c)\sqrt{b-c}$

$=a^3+3a(b-c)-(3a^2+b-c)\sqrt{b-c}$

$=a^3+3a(b-c)$　（因 $3a^2+b-c=0$

$(a-\sqrt{b-c})^3=a^3+3a(b-c)$　………………. (2)

$(a+\sqrt{b-c})^3=(a-\sqrt{b-c})^3$

$a+\sqrt{b-c}=a-\sqrt{b-c}$

$\Rightarrow \sqrt{b-c}=-\sqrt{b-c}$

$\Rightarrow 2\sqrt{b-c}=0$

$\Rightarrow b-c=0$

$\Rightarrow (x+y)^2-4(x^2-xy+y^2)=0$

$\Rightarrow x^2+2xy+y^2-4x^2+4xy-4y^2=0$

$\Rightarrow -3x^2+6xy-3y^2=0$

$\Rightarrow -3(x^2-2xy+y^2)=0$

$\Rightarrow -3(x-y)^2=0$

$\Rightarrow x-y=0$

$\Rightarrow x=y$

Also read

https://johnmayhk.wordpress.com/2010/12/15/not-equal/