A problem by Siddharth Jayshankar

$\sqrt [ 3 ]{ 0.9 } -\sqrt [ 3 ]{ 0.7 } =0.0775\\$

Using ${ (a-b) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab(a-b)$

${ (\sqrt [ 3 ]{ 0.9 } -\sqrt [ 3 ]{ 0.7 } ) }^{ 3 }=0.9-0.7-3\sqrt [ 3 ]{ 0.9\times 0.7 } (\sqrt [ 3 ]{ 0.9 } -\sqrt [ 3 ]{ 0.7 } )\\$ ${ (0.0775) }^{ 3 }=0.2-3(\sqrt [ 3 ]{ 0.63 } )(0.0775)$

$3(\sqrt [ 3 ]{ 0.63 } )=\frac { 0.2-{ (0.0775) }^{ 3 } }{ (0.0775) } =\frac { 0.2 }{ (0.0775) } -{ (0.0775) }^{ 2 }$

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$\sqrt [ 3 ]{ 0.81 } +\sqrt [ 3 ]{ 0.49 } +\sqrt [ 3 ]{ 0.63 } ={ (\sqrt [ 3 ]{ 0.9 } ) }^{ 2 }+{ (\sqrt [ 3 ]{ 0.7 } ) }^{ 2 }+\sqrt [ 3 ]{ 0.9\times 0.7 }$

$Adding\quad and\quad subtracting\quad 2(\sqrt [ 3 ]{ 0.9\times 0.7 )} ,$

${ \{ \sqrt [ 3 ]{ 0.9 } ) }^{ 2 }+{ (\sqrt [ 3 ]{ 0.7 } ) }^{ 2 }-2(\sqrt [ 3 ]{ 0.9\times 0.7 } \} +2(\sqrt [ 3 ]{ 0.9\times 0.7 } )+\sqrt [ 3 ]{ 0.9\times 0.7 }$

$Using\quad { a }^{ 2 }+{ b }^{ 2 }-2ab={ (a-b) }^{ 2 }$

$\{ { (\sqrt [ 3 ]{ 0.9 } -\sqrt [ 3 ]{ 0.7) } ) }^{ 2 }\} +3\sqrt [ 3 ]{ 0.63 } \\$

$By\quad substituting\quad values,\\ { (0.0775) }^{ 2 }+\frac { 0.2 }{ (0.0775) } -{ (0.0775) }^{ 2 }$

$\Rightarrow \sqrt [ 3 ]{ 0.81 } +\sqrt [ 3 ]{ 0.49 } +\sqrt [ 3 ]{ 0.63 } =\frac { 0.2 }{ 0.0775 } \\\Rightarrow \frac { 2 }{ \sqrt [ 3 ]{ 0.81 } +\sqrt [ 3 ]{ 0.49 } +\sqrt [ 3 ]{ 0.63 } } =\frac { 2 }{ \frac { 0.2 }{ 0.0775 } } =2\div (\frac { 0.2 }{ 0.0775 } )=2\times (\frac { 0.0775 }{ 0.2 } )=0.775$

$\therefore \frac { 2 }{ \sqrt [ 3 ]{ 0.81 } +\sqrt [ 3 ]{ 0.49 } +\sqrt [ 3 ]{ 0.63 } } =0.775$