((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.
$${\left({\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$
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
$${{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}$$
$${{\mathtt{2}}}^{{\mathtt{2}}} = {\mathtt{4}}$$
.$${\left({\left({{\mathtt{64}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$
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
$${{\mathtt{2}}}^{\left({\frac{{\mathtt{24}}}{{\mathtt{12}}}}\right)}$$
$${{\mathtt{2}}}^{{\mathtt{2}}} = {\mathtt{4}}$$
By doing it so slowly to show you all the steps it actually made it seem much harder. :)