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) |