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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 (ab)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 (ab)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)(ab)

( 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

Feedback

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.