Algebraic Identities for class 9

Algebraic identities: Algebraic identities are an equation that is true for all values of variables. Here are some popular identities which we use a lot in mathematics.

(a+b)(a+b)=a²+2ab+b²

(a-b)(a-b)=a²-2ab+b²

(a+b)(a-b)=a²-b²

Difference between identity and equation

Identity: Identity is equality(or equation) which is true for all values of variables.

Example : Identity ( a²-b²)=(a-b)(a+b) is true irrespective of value of variables.

whereas 

Equation: An Equation value varies with values of variables.

Example :  x²+4 =0 ,it’s value vary with the variable .Solution of equation depend on x . There are certain values of x for which given equation is true .

Here are few more algebraic identity which we’ll study in this chapter

(i) (a+b+c)²=a²+b²+c²+2ab+2bc+2ac

(ii) (a+b)³= a³+b³+3ab(a+b)

(iii) (a-b)³=a³-b³-3ab(a-b)

(iv) a³+b³= (a+b)(a²-ab+b²)

(v) a³-b³=(a-b)(a²+ab+b²)

(vi) a³+b³+c³= (a+b+c)(a²+b²+c²-ab-bc-ca)

There is one more identity which is called conditional identity because it applies when certain conditions get satisfied.

Conditional identity: A  conditional identity is an identity that is true when the unknown (s) satisfies a certain condition.

If a+b+c=0,then a³+b³+c³=3abc

Let’s Learn to solve problem-based on algebraic identity

Q.1 Expand (3x+4y)²

Solution(i)

We have,

(3x+4y)²=(3x)²+2.3x.4y+(4y)²       [Using :(a+b)(a+b)=a²+2ab+b²]

(3x+4y)² =9x²+24xy+16y²

Q.2 Find the product

(i) (2x+3y)(2x-3y)

(ii)(2x+y)(2x-y)(4x²+y²)

Solution(i)

We have ,

(2x+3y)(2x-3y)=(2x)²-(3y)²           [Using :(a+b)(a-b)=a²-b²)

(2x+3y)(2x-3y)= 4x²-9y²

Solution(ii) we have ,

(2x+y)(2x-y)(4x²+y²)=(2x)²-(y)²(4x²+y²)        [Using :(a+b)(a-b)=a²-b²)

=  (4x²-y²) (4x²+y²)

=  16x^4y^4

Q.3 Without actual multiplication, find the value of 103×97

Solution: we have, 103×97 which we can write as (100+3)(100-3)

= (100)²-(3)²                                                          [Using :(a+b)(a-b)=a²-b²)

=10000-9

=9991

Q.4 If x+y=12 and xy=32 , find the value of x²+y²

Solution: we know (x+y)²=x²+2xy+y²

It is given that x+y=12, xy=32, and    x²+y²=?

(x+y)²=(12)²                     (squaring on both side)

x²+2xy+y² =144

x²+y²+2xy= 144

x²+y²+2.32=144                                              (xy=32)

x²+y²=144-64

x²+y²=80

Q.5 Prove that a²+b²+c²-ab-bc-ca is always non-negative for all values of a,b and c.

Solution:  We can prove this with the help of contradiction

let a²+b²+c²-ab-bc-ca is equal to any negative value say K

∴                      a²+b²+c²-ab-bc-ca =K          where K is any negative integers

Now multiply both sides by 2

2a²+2b²+2c²-2ab-2bc-2ac=2K

a²+ a²+b²+b²+c²+c²-2ab-2bc-2ac=2K

a²+b²-2ab+b²+c²-2bc+a²+c²-2ac=2K

(a-b)²+(b-c)²+(c-a)²= 2K

Now here we have sum of the square of three numbers and we know the square of any number whether it is positive or negative is always positive.

so, Our assumption is wrong

∴      a²+b²+c²-ab-bc-ca is always non-negative for all values of a,b and c

Q.6 If a²+b²+c²=250 and ab+bc+ca=3 ,find the a+b+c.

Solution :  we know (a+b+c)²=a²+b²+c²+2ab+2bc+2ac

(a+b+c)²= a²+b²+c²+2(ab+bc+ca)

substitute the given value in above equation

(a+b+c)²= 250+2(3)

(a+b+c)²=250+6

(a+b+c)²=256

∴                                   a+b+c =16

Q.7 Expand

(i) (2x+3y)³

(ii) (3x-2y)³

Solution(i)  we can expand it with the help of identity (a+b)³= a³+b³+3ab(a+b)

(2x+3y)³= (2x)³+(3y)³+3.2x.3y(2x+3y)

=  8x³+27y³+36x²y+54xy²

similarly we can also expand

(3x-2y)³= (3x)³-(2y)³-3.3x.2y(3x-2y)

= 27x³-8x³-54x²y+36xy²

Q.8 If x+y = 12 and xy=27 find the value of x³+y³

Solution: 

We can solve in two ways first using identity and second finding the separate value of x and y which we will study in the quadratic equation for class 10 but today we will learn to solve it with both ways.

Using identity (a+b)³= a³+b³+3ab(a+b)

since we have x+y = 12

cubing on both side ,

(x+y)³=12³

x³+y³+3xy(x+y)=1728

x³+y³+3.27.12=1728

x³+y³=1728-972

x³+y³=956

Problem based on conditional identity

Q.Without actual calculations, find the value of the following

(i) (-12)³+(7)³+(5)³

(ii) (28)³+(-15)³+(-13)³

Solution: 

(i) On comparing it with LHS of the identity a³+b³+c³=3abc we get,

a=-12  ,b=7   and c=5

a+b+c= -12 +7+5 =0 hence condition is satisfying

    ∴                 a³+b³+c³=3abc

(-12)³+(7)³+(5)³=3.(-12).7.5

=-36.35

= -1260

similarly we can solve (28)³+(-15)³+(-13)³

28-15-13=28-28=0

∴              (28)³+(-15)³+(-13)³=3×28×-15×-13

=16380

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