2.17 Conditional Identities:

 

We have already studied identities in section 2.3 and let us recollect the concept

 

For any value of  x, y and z we can verify that following.

(x+y)(x+z) = x(x+z)+y(x+z)= x2+xz+xy+yz= x2+x(y+z)+yz

 

By substituting suitable values for x, y, z in the above identity, we can arrive at the following identities.

 

No

Formula

Expansion

1

(a+b)2

a2+b2+2ab

2

(a-b)2

a2+b2-2ab

3

(a+b)(a-b)

a2-b2

4

(a+b+c)2

a2+b2+ c2+2ab+2bc+2ca

5

(x+a)(x+b)(x+c)

x3+ x2(a+b+c) +x(ab+bc+ca)+abc

6

(a+b)3

a3+b3+3ab(a+b)

7

(a-b)3

a3-b3-3ab(a-b)

8

(a+b) (a2+b2-ab)

a3+b3

9

(a-b) (a2+b2+ab)

a3-b3

10

(a+b+c)( (a2+b2 +c2-ab-bc-a)

a3+b3 +c3-3abc

 

2.17 Problem 1: If a+b+c = 0 prove that a2/bc+ b2/ca+ c2/ab = 3

 

Solution:

We need to simplify LHS such that we get an expression 3X/X(=3)

Since a+b+c = 0 we have a =-b-c. b= -a-c, c=-a-b

 

No

Step

Explanation

1

LHS= a3/abc+b3/bca +c3/cab

Multiply both numerator and denominator of each of the term in LHS by a, b, c respectively

2

= (a3+b3+c3)/abc

Take abc out as  the common divisor

3

=(a2.a+bb2+cc2)/abc

split powers of a, b, c

4

=[a2(-b-c) + b(-a-c)2+c(-a-b)2]/abc  

substitute   -b-c, (-a-c)2, (-a-b)2 respectively  for a, b2,c2

5

= [a2(-b-c) + b{-(a+c)}2+c{-(a+b)}2]/abc  

 

6

= [- a2b- a2c +b(a2+ c2+2ac)+ c(a2+b2+2ab)]/abc

expand (a+c)2,  (a+b)2

7

=[- a2b- a2c +ba2+ bc2+2abc+ ca2+cb2+2abc)]/abc 

Terms get cancelled(red color)

8

=[ bc2 +cb2+abc+3abc]/abc

Simplify

9

= [bc(c+b+a)+3abc]/abc

Simplify

10

=[bc(0)+3abc]/abc

a+b+c =0

11

= 3abc/abc =3

 

 

In the above example we simplified the equation using the condition a+b+c = 0

 

Definition :

 

The  identity(equation) which is true for all values of variables  subject to  given conditions are called ‘conditional identities’

 

2.17 Problem 2: If a+b+c = 2S prove that a2+b2- c2+2ab/ a2-b2+ c2+2ac = (S-c)/(S-b)

 

Solution:

 

Let us first take the numerator of LHS

 

No

Step

Explanation

1

a2+b2- c2+2ab

Given numerator

2

= (a2+b2+2ab)-c2

Rearrange terms

3

= (a+b)2- c2

This is of the form X2-Y2 =(X+Y)(X-Y)  with X =a+b and Y=c

4

= ((a+b)+c)(a+b)-c))

 

5

=2S(2S-2c)

It is given that a+b+c = 2S and hence a+b-c = a+b+c-2c=2S-2c

6

a2-b2+ c2+2ac =2S(2S-2b)

Follow the above steps to simplify denominator of LHS

 

LHS = a2+b2- c2+2ab/ a2-b2+ c2+2ac

= 2S(2S-2c)/2S(2S-2b)

=2(S-c)/2(S-b)

=(S-c)/(S-b) = RHS

 

2.17 Problem 3: If a+b+c = 2S prove that S2+(S-a)2+ (S-b)2+(S-c)2= a2+b2+c2

 

Solution:

 

No

Step

Explanation

1

LHS = S2 +(S2+a2-2aS)+(S2+b2-2bS)+(S2+c2-2cS)

Expand individual terms using (a+b)2

2

= 4S2+ a2+b2+c2-2S(a+b+c)

Substitute 2S for (a+b+c)

3

=4S2+ a2+b2+c2-2S*2S =4S2+ a2+b2+c2-4S2

 

4

= a2+b2+c2= RHS

 

 

2.17 Problem 4: If a+b+c 0 and  a3+b3+c3=3abc prove that a=b=c

 

Hint:

By using suitable identity, arrive at a condition that

{(a-b)2+ (b-c)2+(c-a)2} =0

If sum of three positive terms have to be zero then it is necessary that each term has to be zero

(a-b)2=0, (b-c)2=0, (c-a)2=0

 a-b= 0, b-c=0,c-a =0

 

 

2.17 Summary of learning

 

 

No

Points studied

1

Solving of conditional identities(S)