((64^(3/4)*(sqrt(2)/2))^(-1/2)*(4sqrt(2))^3*sqrt(2))^(1/3)

I need to solve this, but i simply don't know how to start. So anyone can solve it, and post here every step of it?

Thanks.

Guest Dec 28, 2014

#1**+13 **

$${\left({\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

mmm that looks scary doesn't it. :)

$${\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}}{{{\mathtt{2}}}^{{\mathtt{1}}}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{2}}}^{{\mathtt{6}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{8}}}{{\mathtt{12}}}}\right)}$$

Now -9+1+24+8=24

$${{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}$$

$${{\mathtt{2}}}^{{\mathtt{2}}} = {\mathtt{4}}$$

.Melody Dec 28, 2014

#1**+13 **

Best Answer

$${\left({\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

mmm that looks scary doesn't it. :)

$${\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}}{{{\mathtt{2}}}^{{\mathtt{1}}}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}{\left({{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

$${{\mathtt{64}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

$${\left({{\mathtt{2}}}^{{\mathtt{6}}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{8}}}}\right)}{\mathtt{\,\times\,}}\left({{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}\right){\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}$$

$${{\mathtt{2}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{8}}}{{\mathtt{12}}}}\right)}$$

Now -9+1+24+8=24

$${{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}$$

$${{\mathtt{2}}}^{{\mathtt{2}}} = {\mathtt{4}}$$

Melody Dec 28, 2014

#2**+10 **

By doing it so slowly to show you all the steps it actually made it seem much harder. :)

Melody Dec 28, 2014