Algebraic Identities for class 9

Introduction

we have already studied about algebraic expressions their addition ,subtraction ,multiplication and division .when we multiply binomial(Algebraic Expressions having two terms),we obtained the following results which is algebraic identities because it is true for all values of a and b i.e LHS=RHS

(a+b)(a+b)=a²+2ab+b²

            (a-b)(a-b)=a²-2ab+b²

           (a+b)(a-b)=a²-b²

 

Here are few algebraic identity which we shall 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 condition get satisfied.

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

STATEMENT  If a+b+c=0,then prove that a³+b³+c³=3abc

Difference between identity and equation

Identity: A Identity is equality which is true for all values of variables.

Equation: An Equation values varies with different values of variables.

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)

=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)

= 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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