# Simplification

The process of simplifying a large and complicated expression into a number or fraction is known as simplification. To solve an expression, all the given operation (addition, subtraction, multiplication, division etc) are solved in a particular manner which is known as

**VBODMAS.**

**Rule of VBODMAS**

**VBODMAS**is the sequence of operations in which they are to be solved.

Meaning of
each letter of

**VBODMAS**is as follow:- V – Viniculum or Bar (-)
- B – Brackets ((), {}, [])
- O – Of
- D – Division
(÷)
- M –
Multiplication (×)
- A – Addition (+)
- S – Subtraction (-)

**Note**that while solving brackets, first solve small brackets

**(),**then curly brackets

**{}**, and at last square brackets

**[]**.

## Algebraic Formula

Algebraic
formula are useful in simplifying expression.

Important Formulas in Algebra

**Here is a list of Algebraic formulas**–

- a
^{2}– b^{2}= (a – b) (a + b)- (a+b)
^{2}= a^{2}+ 2ab + b^{2}- a
^{2}+ b^{2}= (a – b)^{2}+ 2ab- (a – b)
^{2}= a^{2}– 2ab + b^{2}- (a + b + c)
^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2ac + 2bc- (a – b – c)
^{2}= a^{2}+ b^{2}+ c^{2}– 2ab – 2ac + 2bc- (a + b)
^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3}; (a + b)^{3}= a^{3}+ b^{3}+ 3ab(a + b)- (a – b)
^{3}= a^{3}– 3a^{2}b + 3ab^{2}– b^{3}- a
^{3}– b^{3}= (a – b) (a^{2}+ ab + b^{2})- a
^{3}+ b^{3}= (a + b) (a^{2}– ab + b^{2})- (a + b)
^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3}- (a – b)
^{3}= a^{3}– 3a^{2}b + 3ab^{2}– b^{3}- (a + b)
^{4}= a^{4}+ 4a^{3}b + 6a^{2}b^{2}+ 4ab^{3}+ b^{4})- (a – b)
^{4}= a^{4}– 4a^{3}b + 6a^{2}b^{2}– 4ab^{3}+ b^{4})- a
^{4}– b^{4}= (a – b) (a + b) (a^{2}+ b^{2})- a
^{5}– b^{5}= (a – b) (a^{4}+ a^{3}b + a^{2}b^{2}+ ab^{3}+ b^{4})If n is a natural number, a^{n}– b^{n}= (a – b) (a^{n-1}+ a^{n-2}b+…+ b^{n-2}a + b^{n-1})If n is even(n = 2k), a^{n}+ b^{n}= (a + b) (a^{n-1}– a^{n-2}b +…+ b^{n-2}a – b^{n-1})If n is odd(n = 2k + 1), a^{n}+ b^{n}= (a + b) (a^{n-1}– a^{n-2}b +…- b^{n-2}a + b^{n-1})- (a + b + c + …)
^{2}= a^{2}+ b^{2}+ c^{2}+ … + 2(ab + ac + bc + ….

###
**Simplification Tricks**

As we know that

**takes more time to solve the problem. If you are doing preparation of government exam then many questions have asked form simplification. In competitive examination we have limited time and we have to solve many problems.***simplification*
Therefore, here I will give you few

**which help you to solve the problem and I’m sure after these tricks you are able to solve problem within second.***simplification tricks***Simplification Trick No 1.**

If given expression in this form

Then, final result will be

**a + b**

**Example: Solve the below given equation:-**

Here,

**a = 0.05**and**b = 0.04**

Therefor, Answer will be

**a + b**= 0.05 + 0.04 =**0.09 Ans.**

**Simplification Tricks No 2.**

If given expression in this form

Then final result will be

**a - b**

**Example:**

Therefor, Answer will be

**a - b**= 0.05 – 0.04 =**0.01 Ans**.**Simplification Trick No 3.**

If given expression in this form

Then, Answer will be

**a + b + c**

**Example:**

**Ans: -**

**a + b + c**= 325 + 425 + 625 =

**1375**

**Simplification Trick No 4.**

If given expression in this form

**2**

**Example:**

**2**Ans.

**Note:- In place of a & b any value given, ans always be 2.**

**Simplification Trick No 5.**

If given expression in this form

Then, Answer will be

**4.**

**Example:**

When you will solve this problem the
you will get

**4 Ans.**

**Simplification Trick No. 6**

If given expression in this form

Therefor, Answer will be

**{a + b}**

**Example:**

Here,

**a = 9 & b = 8****Ans = a + b = 9 + 8 = 17**

**Simplification Trick No. 7**

If given expression in this form

**Therefore, Answer will be**

**Example:**

**Solution Here,**

**Simplification Trick No 8.**

If sum of two numbers and product of
two numbers are given then difference of two number will be

**Example:**

**If Sum of two number is 10 & product of those number is 2 then what is difference between two numbers.**

Here,

**X + Y**= 10 &**XY**= 2
Then, By putting this value in above equation we get

**X - Y**= Root 2.##
**Conclusion**

This all simplification tricks will help you in Competitive examination. You just remember the Form of equation. If you have remember the form of equation then you can solve the problem within sec.

I Hope you like all shortcut trick.

I will upload more shortcut trick on my website so whenever you need this shortcut trick you can visit on my website and in MATH section you will get all Math problem.

If you have any problem regarding this so please write your problem in the comment box. I will happy to help you.

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