<PRE><B>Question 1032410</B>
The expression sqrt(729g^5y^(1/3))^3 (729g^5y^3) equals kg^ry^s
-------
= 729^(3/2)*g^(15/2)*(y))(729*g^5*y^3)  = k*g^r*y^s
-----
= 729^(5/2)*g^(25/2)*y^(6)
------------------- 
where r, the exponent of g, is: r = 0
---------------------------------------- 
And s, the exponent of y is: 1+3 = 4
-----------------------------------
And k, the leading coefficient is: (3/2)+1 = 5/2
---------------------------
The exponent raised to the 3 on the first squate root is supposed to be a cube(third) root
------
Cheers,
Stan H.
-------------</PRE>