Anjan Debnath

Dear Roshan.
I have seen your query and as a maths professor, am trying my best to answer your doubt step by step, from my point of view and experience as well.
 
First of all, as an instructor in mathematics,o i must advise you not to pay attention to neon thinking about this kind of questions. Your first and most essential nature should be to prepare yourself for the board exam with a bright concert in the subject matter. If you think about "how many questions might appear in exam outside the books" believe me or not, at the end of the day, somehow this will affect your mind, will make you frightened about the exam and eventually will change in your result. That's not the end. You might lose confidence in our preparation already you made. It means the portion you have studied and know that there will be a question from the part you studied, will become confusing after a while.
 
It is 100% guaranteed that a few questions will arrive in the exam paper which you will never find in the book. But what is important to notice or observe is that these questions will be entirely conceptual. On other words, you will be able to solve them only if your concept in subject matter is clear. Let me give you an example. 
 
Say A is a matrix of order 3x3 given by 
( [ 1,2,3 ], [0,2,3] , [ 0,0,3] )
Where I have written each row elements within square brackets. Means, 1,2,3 are the elements in the first row, 0,2,3 are the elements in second row etc. Suppose you are asked to find the determinant of A^4. 
 
What will you do? Are you tempted to say, sir, I will find A^4 by using Axa and then A^2 x A^2 and then find the determinant ?! Well, that will do the job but imagine if they ask you the same question but for A^100. Then? You are not going to take the above approach, are you? So how to solve this one? 
 
Simple. Observe that A is an upper triangular matrix and the determinant of the upper triangular matrix the product of the diagonal elements. In that way, determinant of A is 1x2x3, i.e. 6. 
 
Now det(A^4)={ det(A) }^4. Hence the required answer will be 6^4. As you can see, the determinant of A^100 has now become easier as well.
 
The above question naturally clear that, you can tackle such question through a clear concept only. These kinds of questions will appear as question outside NCERT math book, and if your preparation and concepts are strong enough, you don't have to worry about the exam paper any more.
 
This is why I am advising you. Study the entire book. First, finish every exercise. Once you are done, move into the examples which are provided in every chapter just before the use. Study them. Sometimes question appears from this zone also, which most of the time, students do not bother to study and let me share a tip, several conceptual issues you can find are hidden here. 
 
I hope I have addressed your query. But if you still have any doubt, let me know. 
 
I wish you a bright future ahead.
 
Take care!
 
 
 
 
 
 
 
 
 
 

Yes, there are. I am trying to explain step by step.
Say we want the square of n = 67. First of all, the square of 7 is 49. Write 9 in the answer place and keep 4 for carrying over process.
∴n^2 = 9
Then, see that 6 and 7 form the number 67. Their product is 42. The rule says you have to double it. Hence you get double of 42 as 84. Also, you have 4 in your hand for carrying over. Together you get 84+4=88. Write the unit digit eight beside 9 and keep 10s digit 8 for next round carry-over process.
∴ n^2 = 89
Finally, the 10s digit is 6. Square it, and you will get 36. Plus you have 8 in your hand. Together we get 36+8, i.e. 44. Now write it beside 89, and we are done.
∴ 67^2 = 4489.
Try your self: 39^2, 23^2, 52^2.
 
Verify your answer using a calculator.
 
For squaring three digits or more, there is procedure indeed, but I can't explain them here within short place. 
 
Although I can show you some other special technique for squaring 2 digit numbers of special types. 
 
1. If the number ends with 5: 
Let's say we need to square 85. First, write 25 straight forward in answer place. 
85^2 = 25
 
Then, note that after ignoring five we are left with the number 8 in the given number 85. The next number of 8 is 9. The rule says to make the product which is in our present case is 8 x 9, i.e. 72.
 
Write it down, and we are done.
85^2 = 7225.
Similarly, 95^2 = 9025 because 9 x 10 = 90 and 25 already is written etc.
115^2 = 13225 because 11 x 12 = 132. Etc
 
2. If the number is near to 100.
Let's say we need to square 93. 
 
The difference between 93 and 100 is 7. The rule says to write down the square of 7 first, which is here 49.
Next, decrease 93 by 7. Which gives 86. Hence the final answer becomes 
93^2=8649.
Similarly, 95^2 is 9025 because the difference between 95 and 100 is 5. The square of 5 is 25, and if we decrease 95 by 5, we shall get 90
 Together we thus obtain 
95^2= 9025.
Notice we have achieved the same already before by using the general squaring technique explained above.
 
Caution. Suppose we want to square 98. This time doesn't be tempted to write 964. That's not correct. While you are squaring the difference between 100 and the given number, make sure the answer occupies exactly two places only. If not, use 0 to fill up the lattice. Hence 98^2 is not 964 but in 9604.
 
Similarly, for 88^2, first, write down the square of the difference which is 12^2, i.e. 144. But this time it is a three-digit answer. So keep the last two digits and carry over the excess digit which is 1 in this present case. Now decrease 88 by 12, and we get 76. Added by carried over one we get 77. Hence 88^2 = 7744.
 
There are more so many special rules. But we can discuss them some other days.
 
Take care.