How do i work this out? 10pts?

http://img509.imageshack.us/my.php?image鈥?/a>How do i work this out? 10pts?
For the square root of n cubed to be an integer, n must be a perfect square. So I suggest that we replace n with N^2, so we can simplify out the square root sign.





sqrt(n^3) = sqrt((N^2)^3) = sqrt(N^6) = N^3





Now we need number of all integers N such that 0 %26lt; N^3 %26lt; 200. To solve, we simply take the cube root of all sides.





0 %26lt; N %26lt; cbrt(200) = 5.85 (2 dp)





We take the lower integer to make sure we are in the boundaries:





1 %26lt;= N %26lt;= 5





So we have 5 numbers (one for each 1, 2, 3 ,4, 5). Each of those numbers will correspond with the original n, which will produce the results: (1, 4, 9, 16, 25), but still 5 is your answer.How do i work this out? 10pts?
For a positive integer n


(n^3)^(1/2) %26lt;= 200


equals


n %26lt;= (200^2)^(1/3) =34.1995


so there are at least 34 different integers (from 1 to 34) that fullfill the equation. Anything bigger fails the equation - so D is the correct answer.
34.


Plug each of those answers into the equation.


One of them will be just under 200.


Every value of n below that will satisfy the equation.


in this case, when u plu 34 in, you get 198 or 199.


if u plug in 35 u get over 200.


Any number lower than 34 u will get less than 198.
n^(3/2) %26lt; 200


Or n^3 %26lt; 200^2 = 40000 %26gt; 39304 = 34^3


Or n = 1, 2, 3, ...., 31, 32, 33, 34.


n has 34 values.
I have no idea! Sorry...