finding the square, cube and nth power of the number which contains only 9s

• (Number containing 9s)^2 - first write N-1 9s where N is the no of digits in the number, followed by 8 and then followed by N-1 0s, followed by 1

EX: 99^2 = 9801
there are 2 9s so N=2 and N-1 is 1, hence first digit will be 9 followed by 8 and again N-1 zeros and then 1 which is 9801

llly 999^2 = 998001

• (Number containing 9s)^3 - first write N-1 9s where N is the no of digits in the number, followed by 7 and then followed by N-1 0s, followed by 2 and finally N 9s. 

EX: 99^3 = 970299
here you can see only 2 9s so N will be 2 and N-1 wil be 1, hence first digit will be 9 followed by 7 and followed by N-1 0s followed by 2 and finally N 9s

llly 999^3 = 997002999

• (Number containing 9s)^4 - first write N-1 9s where N is the no of digits in the number, followed by 6 and then followed by N-1 0s, followed by 5, again followed by N-1 9s, followed by 6 and N-1 0s and finally 1

EX: 99^4 = 96059601
here you can see only 2 9s so N will be 2 and N-1 wil be 1, hence first digit will be 9 followed by 6 and followed by N-1 0s followed by 5 and then again N-1 9s followed by 6 and N-1 0s finally 1

llly 999^4 = 996005996001

This is how we can easliy find out the nth root of number containing only 9s

To summarize:
for square - remember 8,1 in mind
for cube - remember 7,2
for 4th root - remember  6,5,6