[6666.....6 upto n digits] whole square + [8888.....8 upto n digits ] = ?

Let's first perform some trials with n=1,2 and 3.When n=1,then, Given expression yields the value 6^2 + 8^2 = 10^2.Similarly, When n=2 ,It yields the value (66)^2 + (88)^2 = (11*6)^2 + (11*8)^2 = (11^2)*(6^2 + 8^2) = 11^2 * 10^2 = 110^2.Similarly, For n = 3, we get its value as (666)^2 + (888)^2 = (111)^2 *(10^2) = (1110)^2 .So,Here,we conclude that a series of values are forming at the succesive natural values of n which can be represented as the square of a number which is the product of 10 and a number containing a total of 'n' numbers of digit 1 (and nothing else).so, (6666...upto n digits)^2 + (8888...upto n digits)^2 = [1111....(upto n times)]*(100) = [1111...(upto n times)0]^2 .[Sign n^2 means n square.]