If N=9^9, then N is divisible by how many positiveperfect cubes

1.	If N=9^9, then N is divisible by how many positiveperfect cubes
Answer» $Given \:N = 9^{9}$ $\implies N = ( 3^{2} )^{9}$ $\implies N = ( 3)^{2\times 9}$ /* By EXPONENTIAL LAW / $\boxed{\pink { (a^{m})^{n} = a^{m\times n } }}$ $\implies N = ( 3 )^{18}$ $\implies N = ( 3 )^{6\times 3}$ $\implies N = (3^{6})^{3}$ $\red{ So, Cubes \:will \:be} = 6 + 1$ ________________ / We KNOW that */ $Number\:of \:factors \:of \:a\: number$ $x = a^{p}\times b^{r}\times c^{s} \ldots \:is$ $( p + 1 )( q + 1 ) ( s + 1 ) \ldots$ _________________ $\green {= 7}$ THEREFORE., $\red{ Number \:of \: positive \: perfect \:cubes}$ $\green { = 7}$ •••♪

Answer»

$Given \:N = 9^{9}$