Given Equation:-
![\sf25 - 12 \div [(8 - 5) \times 1]](https://tex.z-dn.net/?f=%20%5Csf25%20-%2012%20%5Cdiv%20%5B%288%20-%205%29%20%5Ctimes%201%5D)

Step-by-step explanation:-

![\dashrightarrow \sf25 - 12 \div [(8 - 5) \times 1]](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%20%5Csf25%20-%2012%20%5Cdiv%20%5B%288%20-%205%29%20%5Ctimes%201%5D)
first write the equation so that you'll emit minor mistakes of writing wrong digits.

![\dashrightarrow \sf25 - 12 \div [(3) \times 1]](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%20%5Csf25%20-%2012%20%5Cdiv%20%5B%283%29%20%5Ctimes%201%5D)
subtract 8 and 5 , which will give us result as 3

![\dashrightarrow \sf25 - 12 \div [3 \times 1]](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%20%5Csf25%20-%2012%20%5Cdiv%20%5B3%20%5Ctimes%201%5D)
open the brakets that contains 3.

![\dashrightarrow \sf25 - 12 \div [3]](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%20%5Csf25%20-%2012%20%5Cdiv%20%5B3%5D)
multiply 3 with 1 which will result 3.


open the brakets of 3.


to make it easy to understand I have converted it into fraction form.


simplify 12 which is contained in numerator.


as we can see after converting into fraction 3 is common both in numerator and denominator. So cancel ir.


we no longer need the fraction as 3 is canceled.


multiply 2 with 2 which will result 4.


finally after subtracting 25 and 4 , we got our answer as 21.

Rule applied:-
B : bracket
O : on/off
D: divide
M : multiply
A : add
S : subtract