Prove this hardest thing

**jashansinghal** · June 25th 2009, 07:13 PM

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

**Soroban** · June 25th 2009, 08:50 PM

Hello, jashansinghal!

Wow . . . This is a messy one!

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

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

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

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

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

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

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

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

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

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

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

**malaygoel** · July 7th 2009, 08:40 PM

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

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

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

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

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

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

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

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

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