Author: S.LAL
Created: 03 Sep, 2010; Last Modified: 29 Apr, 2018
Algebra II - 04
Expansions
Squares of binomials: (a±b)2
A binomial is an algebraic expression having two terms. If a and b are real numbers or algebraic expressions, then
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
.......(1)
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
.......(2)
Tip: Eq. is enough to evaluate the case for (a−b)2: you just have to replace b
by -b
in the identity and its expansion.
The above expansions of
(
a
+
b
)
2
and
(
a
−
b
)
2
can also be expressed succinctly as:
(
a
±
b
)
2
=
a
2
±
2
a
b
+
b
2
Some other helpful identities that can be derived from Eq. and Eq. are
(
a
+
b
)
2
+
(
a
−
b
)
2
=
2
(
a
2
+
b
2
)
.......(3)
(
a
+
b
)
2
−
(
a
−
b
)
2
=
4
a
b
.......(4)
Since
(
a
+
1
a
)
2
=
a
2
+
2
+
1
a
2
, therefore
a
2
+
1
a
2
=
(
a+
1
a
)
2
−2
.......(5)
Since
(
a
−
1
a
)
2
=
a
2
−
2
+
1
a
2
, therefore
a
2
−
1
a
2
=
(
a
−
1
a
)
2
+
2
.......(6)
Square of trinomial: (a+b+c)2
A trinomial is an algebraic expression having three terms. If a, b and c are real numbers or algebraic expressions, then
(
a
+
b
+
c
)
2
=
(
a
+
(
b
+
c
)
)
2
=
a
2
+
2
a
(
b
+
c
)
+
(
b
+
c
)
2
=
a
2
+
2
a
b
+
2
a
c
+
b
2
+
2
b
c
+
c
2
=
a
2
+
b
2
+
c
2
+
2
a
b
+
2
b
c
+
2
c
a
=
a
2
+
b
2
+
c
2
+
2
(
a
b
+
b
c
+
c
a
)
Thus,
(
a
+
b
+
c
)
2
=
a
2
+
b
2
+
c
2
+
2
(
a
b
+
b
c
+
c
a
)
.......(7)
If
a
2
+
b
2
+
c
2
=
29
and
a
+
b
+
c
=
9
, find
a
b
+
b
c
+
c
a
. [(Bansal, Mathematics Class IX, p. 46, Ex. 3)]
Since
(
a
+
b
+
c
)
2
=
a
2
+
b
2
+
c
2
+
2
(
a
b
+
b
c
+
c
a
)
therefore, we have
9
2
=
29
+
2
(
a
b
+
b
c
+
c
a
)
⇒
81
−
29
2
=
a
b
+
b
c
+
c
a
⇒
a
b
+
b
c
+
c
a
=
26
If
x
+
1
x
=
2
, find (i)
x
2
+
1
x
2
(ii)
x
4
+
1
x
4
. [(Bansal, Mathematics Class IX, p. 46, Ex. 4)]
(i) We know that
x
2
+
1
x
2
=
(
x
+
1
x
)
2
−
2
thus,
x
2
+
1
x
2
=
2
2
−
2
=
2
(ii) Extrapolating from the identity
x
2
+
1
x
2
=
(
x
+
1
x
)
2
−
2
, we have
x
4
+
1
x
4
=
(
x
2
+
1
x
2
)
2
−
2
	
=
(
(
x
+
1
x
)
2
−
2
)
2
−
2
thus,
x
4
+
1
x
4
=
(
2
2
−
2
)
2
−
2
=
2
Cubes of binomials: (a±b)3
If a and b are real numbers or algebraic expressions, then
(
a
+
b
)
3
=
(
a
+
b
)
(
a
+
b
)
2
=
(
a
+
b
)
(
a
2
+
2
a
b
+
b
2
)
=
a
3
+
2
a
2
b
+
a
b
2
+
a
2
b
+
2
a
b
2
+
b
3
=
a
3
+
3
a
2
b
+
3
a
b
2
+
b
3
Thus,
(
a
+
b
)
3
=
a
3
+
3
a
2
b
+
3
a
b
2
+
b
3
.......(8)
Tip: Eq. is enough to evaluate the case for (a−b)3: you just have to replace b
by -b
in the identity and its expansion.
Also,
(
a
−
b
)
3
=
(
a
−
b
)
(
a
−
b
)
2
=
(
a
−
b
)
(
a
2
−
2
a
b
+
b
2
)
=
a
3
−
2
a
2
b
+
a
b
2
−
a
2
b
+
2
a
b
2
−
b
3
=
a
3
−
3
a
2
b
+
3
a
b
2
−
b
3
(
a
−
b
)
3
=
a
3
−
3
a
2
b
+
3
a
b
2
−
b
3
.......(9)
The above expansions of
(
a
+
b
)
3
and
(
a
−
b
)
3
can also be expressed succinctly as:
(
a
±
b
)
3
=
a
3
±
3
a
2
b
+
3
a
b
2
±
b
3
Some other helpful identities that can be derived from Eq. and Eq. :
a
3
+
b
3
=
(
a
+
b
)
3
−
3
a
b
(
a
+
b
)
.......(10)
a
3
−
b
3
=
(
a
−
b
)
3
+
3
a
b
(
a
−
b
)
.......(11)
Tip: Eq. and Eq. can be got by replacing b with
1
a
in Eq. and Eq. respectively.
a
3
+
1
a
3
=
(
a
+
1
a
)
3
−
3
(
a
+
1
a
)
.......(12)
a
3
−
1
a
3
=
(
a−
1
a
)
3
+3(
a−
1
a
)
.......(13)
Write the following cubes in expanded form (i)
(
2
p
−
9
q
)
3
(ii)
(
−
3
x
+
5
y
)
3
. [(NCERT, Mathematics Class IX, p. 100 Example 10(ii); 11(i))]
(i) Comparing with Eq. , we have a =
2p
and b =
−
9q
(
2
p
−
9
q
)
3
=
(
2
p
)
3
+
3
(
2
p
)
2
(
−
9
q
)
+
3
(
2
p
)
(
−
9
q
)
2
+
(
−
9
q
)
3
=
8
p
3
−
108
p
2
q
+
486
p
q
2
−
729
q
3
(i) Comparing with Eq. , we have a =
−
3x
and b =
5y
(
−
3
x
+
5
y
)
3
=
(
−
3
x
)
3
+
3
(
−
3
x
)
2
(
5
y
)
+
3
(
−
3
x
)
(
5
y
)
2
+
(
5
y
)
3
=
−
27
x
3
+
135
x
2
y
−
225
x
y
2
+
125
y
3
Evaluate
1001
3
by making use of suitable identity. [(NCERT, Mathematics Class IX, p. 101 Example 14)]
Writing 1001
as (1000
+
1
) and using Eq. ,
1001
3
=
(
1000
+
1
)
3
=
1000
3
+
3
(
1000
)
2
(
1
)
+
3
(
1000
)
(
1
)
2
+
1
3
=
1000000000
+
3000000
+
3000
+
1
=
1003003001
Multiplication of two binomials: (x±a)(x±b)
If a and b are real numbers or algebraic expressions, then,
(
x
+
a
)
(
x
+
b
)
=
x
2
+
b
x
+
a
x
+
a
b
=
x
2
+
(
a
+
b
)
x
+
a
b
So, we have the identity
(
x
+
a
)
(
x
+
b
)
=
x
2
+
(
a
+
b
)
x
+
a
b
.......(14)
For the additional cases of
(
x
+
a
)
(
x
−
b
)
,
(
x
−
a
)
(
x
+
b
)
and
(
x
−
a
)
(
x
−
b
)
, the same identity of Eq. can be used, with the negative values of a and/or b inserted in the appropriate places.
Evaluate
104
×
106
without directly multiplying the numbers. [(NCERT Mathematics Class IX, p. 90 Example 3)]
Writing 104
as (100
+
4
) and 106
as (100
+
6
)
, and using Eq. ,
(
100
+
4
)
×
(
100
+
6
)
=
100
2
+
(
4
+
6
)
100
+
(
4
×
6
)
=
10000
+
1000
+
24
=
11024
Simplification of (a+b)(a−b)
(
a
+
b
)
(
a
−
b
)
=
a
2
−
a
b
+
b
a
−
b
2
=
a
2
−
b
2
So, we have the identity
(
a
+
b
)
(
a
−
b
)
=
a
2
−
b
2
.......(15)
If
a
b
=
b
c
, prove that
(
a
+
b
+
c
)
(
a
−
b
+
c
)
=
a
2
+
b
2
+
c
2
. [ICSE 1993]
TPT
(
a
+
b
+
c
)
(
a
−
b
+
c
)
=
a
2
+
b
2
+
c
2
We have,
LHS
≡
(
a
+
b
+
c
)
(
a
−
b
+
c
)
⇔
(
(
a
+
c
)
+
b
)
(
(
a
+
c
)
−
b
)
⇔
(
a
+
c
)
2
−
b
2
⇔
a
2
+
2
a
c
+
c
2
−
b
2
...
(1)
However, it is given that
a
b
=
b
c
⇒
a
c
=
b
2
Substituting for ac in (1), we get
(
a
+
b
+
c
)
(
a
−
b
+
c
)
⇔
a
2
+
2
b
2
+
c
2
−
b
2
⇔
a
2
+
b
2
+
c
2
≡
RHS
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List of References
Bansal, RK, Concise Mathematics I.C.S.E., Part I – Class IX, New Delhi: Selina Publishers, 2005.
NCERT, Mathematics – Textbook for Class IX, viewed 25 February, 2008, <http://www.ncert.nic.in/textbooks/testing/index.htm>, (n.d.).
Bibliography
Bansal, RK, Concise Mathematics I.C.S.E., Part I – Class IX, New Delhi: Selina Publishers, 2005.
Gupta, SD & Banerjee, A, ICSE Mathematics for Class 9, Patna, India: Bharati Bhawan, 2003.