2 of them are shown below:

**First Quick Operation:**8

^{2}- 7

^{2}= (8 + 7)

**(8 - 7)**= 15 x

**1**= 15

We have applied the special product of a

^{2}- b

^{2}= (a + b)(a - b), to quickly solve the math operation.

Here the quick operation can be achieved because the consecutive numbers when subtracted become ONE. Thus only the other (a + b) need to be computed only!

Try this: 14

^{2}- 13

^{2}

Answer is (14 + 13) =

**27**.

We need not go to the extent of computing 14 x 14 and 13 x 13 to solve the math question. Simple isn't it ?

**Second Quick Operation:**Given (1 / 2) - (1/3), what is the answer?

Answer is, in the conventional solution, (3 - 2) / (2 x 3) = 1 / 6.

But looking at the detailed step, we noticed that the numerator is always equal to ONE.

This is so because the numbers involved are consecutive and thus subtract to "one".

Therefore, fractions involving consecutive numbers as their denominator can be solved easily.

Try this: ( 1/ 14) - (1/15)

Answer is

**1**/ (14 x 15) or 1 /210.

Math is flexible in nature. A complicated arithmetic or algebra questions can be easily solved when we are able to see the "loopholes" within them. The above 2 methods demonstrated this ability.

Have fun exploring math ! Maths Is Interesting! :D

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