Thread: Prove this hardest thing

Prove that--
{(3+2*5^1/4) ÷ (3-2*5^1/4)}^1/4=( 5^1/4 +1) ÷ ( 5^1/4 -1)
If you cannot understand it ,can u teach me how to write it using mathematical signs.
But for the time being prove it......
It is simply impossible to prove it
I have spent my 1 month on it
but i assure you that this question is correct

Hello, jashansinghal!

Wow . . . This is a messy one!

Prove that: .$\displaystyle \left(\frac{3+2\!\cdot\!5^{\frac{1}{4}}} {3 - 2\!\cdot\!5^{\frac{1}{4}}}\right)^{\frac{1}{4}} \;=\; \frac{5^{\frac{1}{4}} +1} {5^{\frac{1}{4}} -1}$

This is equivalent to: .$\displaystyle \frac{3 + 2\!\cdot\!5^{\frac{1}{4}}} {3 - 2\!\cdot\!5^{\frac{1}{4}}} \;=\;\left(\frac{5^{\frac{1}{4}} + 1} {5^{\frac{1}{4}} - 1}\right)^4 $ $\displaystyle = \;\frac{\left(5^{\frac{1}{4}}+1\right)^4}{\left(5^ {\frac{1}{4}} - 1\right)^4} $



On the right side, the numerator is: .$\displaystyle \left(5^{\frac{1}{4}}\right)^4 + 4\left(5^{\frac{1}{4}}\right)^3 + 6\left(5^{\frac{1}{4}}\right)^2 + 4\left(5^{\frac{1}{4}}\right) + 1 $

. . . . . $\displaystyle = \;\;5 + 4\cdot5^{\frac{3}{4}} + 6\cdot5^{\frac{1}{2}} + 4\cdot5^{\frac{1}{4}} + 1 \;\;=\;\;6 + 6\cdot5^{\frac{1}{2}} + 4\cdot5^{\frac{3}{4}} + 4\cdot5^{\frac{1}{4}} $

Factor: .$\displaystyle 6\left(1 + 5^{\frac{1}{2}}\right) + 4\cdot5^{\frac{1}{4}}\left(5^{\frac{1}{2}}+1\right ) \;=\;\left(6+4\cdot5^{\frac{1}{4}}\right)\left(5^{ \frac{1}{2}} + 1\right)$ $\displaystyle = \;2\left(3+2\cdot5^{\frac{1}{4}}\right)\left(5^{\f rac{1}{2}}+1\right) $



On the right side, the denominator is: .$\displaystyle \left(5^{\frac{1}{4}}\right)^4 - 4\left(5^{\frac{1}{4}}\right)^3 + 6\left(5^{\frac{1}{4}}\right)^2 - 4\left(5^{\frac{1}{4}}\right) + 1 $

. . . . . $\displaystyle = \;\;5 - 4\cdot5^{\frac{3}{4}} + 6\cdot5^{\frac{1}{2}} - 4\cdot5^{\frac{1}{4}} + 1 \;\;=\;\;6 + 6\cdot5^{\frac{1}{2}} - 4\cdot5^{\frac{3}{4}} - 4\cdot5^{\frac{1}{4}} $

Factor: .$\displaystyle 6\left(1 + 5^{\frac{1}{2}}\right) - 4\cdot5^{\frac{1}{4}}\left(5^{\frac{1}{2}}+1\right ) \;=\;\left(6-4\cdot5^{\frac{1}{4}}\right)\left(5^{\frac{1}{2}} + 1\right)$ $\displaystyle = \;2\left(3-2\cdot5^{\frac{1}{4}}\right)\left(5^{\frac{1}{2}}+ 1\right) $



The fraction on the right becomes: .$\displaystyle \frac{2\left(3+2\cdot5^{\frac{1}{4}}\right)\left(5 ^{\frac{1}{2}}+1\right)} {2\left(3-2\cdot5^{\frac{1}{4}}\right)\left(5^{\frac{1}{2}}+ 1\right)}$ $\displaystyle = \;\frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}} $


Hence, we have: .$\displaystyle \left(\frac{5^{\frac{1}{4}}+1}{5^{\frac{1}{4}}-1}\right)^4 \;=\; \frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}}$


Therefore: .$\displaystyle \frac{5^{\frac{1}{4}}+1}{5^{\frac{1}{4}}-1} \;=\;\left(\frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}}\right)^{\frac{1}{4}} $

Let $\displaystyle y=\frac{(x+1)^4}{(x-1)^4}$

$\displaystyle \frac{y+1}{y-1}=\frac{(x+1)^4+(x-1)^4}{(x+1)^4-(x-1)^4}$

$\displaystyle \frac{y+1}{y-1}=\frac{x^4+6x^2+1}{4x^3+4x}$

substitute $\displaystyle x=5^{\frac{1}{4}}$

$\displaystyle \frac{y+1}{y-1}=\frac{6+6x^2}{4x^3+4x}$

$\displaystyle \frac{y+1}{y-1}=\frac{6}{4x}$

$\displaystyle \frac{y+1}{y-1}=\frac{3}{2\cdot5^{\frac{1}{4}}}$

$\displaystyle y=\frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}}$