Download expansions - I Love Maths

algebra
Expansions
13
Introduction
In the previous classes, you have studied about algebraic expressions and the basic operations
on them i.e. their addition, subtraction, multiplication and division. You have also studied
some special products and expansions. In this chapter, we shall study some more special
products and expansions, and will learn their applications.
3.1 Special Products
The following products occur frequently in algebra, and the students are advised to memorize
them (without resorting to actual multiplication) until they can recognise each, both the
product from the factors and the factors from the product.
1. (a + b)2 = a2 + 2 ab + b2.
2. (a – b)2 = a2 – 2 ab + b2.
3. (a + b) (a – b) = a2 – b2.
1 2
1


4.  a +  = a2 + 2 + 2.
a
a
1 2
1
5.  a −  = a2 + 2 – 2.

a
a
1
1
1
6.  a +   a −  = a2 − 2 .

a 
a
a
7. (i) (x + a) (x + b) = x 2 + (a + b) x + ab.
(ii) (x + a) (x – b) = x 2 + (a – b) x – ab.
(iii) (x – a) (x + b) = x 2 – (a – b) x – ab.
(iv) (x – a) (x – b) = x 2 – (a + b) x + ab.
8. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).
9. (a + b) 3 = a 3 + b 3 + 3 ab (a + b) = a 3 + 3 a 2 b + 3 ab 2 + b 3.
10. (a – b) 3 = a 3 – b 3 – 3 ab (a – b) = a 3 – 3 a 2 b + 3 ab 2 – b 3.
11. (a + b) (a 2 – ab + b 2) = a 3 + b 3.
12. (a – b) (a 2 + ab + b 2) = a 3 – b 3.
13. (a + b + c) (a 2 + b 2 + c 2 – ab – bc – ca) = a 3 + b 3 + c 3 – 3abc.
14. (x + a) (x + b) (x + c) = x 3 + (a + b + c) x 2 + (ab + bc + ca) x + abc.
Remarks
❏ The expression a2 + 2 ab + b2 is called the expansion of (a + b) 2, a 2 – 2 ab + b 2 is the expansion
of (a – b) 2 and a 3 + b 3 + 3 ab (a + b) is the expansion of (a + b) 3 and so on.
❏ The above results 1 to 3 and 7 to 14 are true for all values of the variables involved, and
the results 4 to 6 are true for all values of a except 0. An equation which is true for all values of
the variables involved is called an identity. Thus, the above results 1 to 14 are all identities ; of
course, a ≠ 0 in the results 4 to 6.
3.2 Applications of the special products
We learn the applications of the special products with the help of the following examples :
Illustrative Examples
Example 1. Find the expansions of the following:
(i) (2 a + 3 b) 2
(ii) (3 x – 4 y) 2
(iii) (2 a + 3 b – 4 c) 2.
Solution. (i) (2 a + 3 b) 2 = (2 a) 2 + 2(2 a) (3 b) + (3 b) 2
= 4 a 2 + 12 ab + 9 b 2.
(3x – 4y)2 = (3 x) 2 – 2(3 x) (4 y) + (4 y) 2
(ii)
= 9 x 2 – 24 xy + 16 y 2.
(2a + 3b – 4c)2 = [2a + 3b + (– 4c)]2
(iii)
= (2 a) 2 + (3 b) 2 + (– 4 c) 2 + 2 [(2 a) (3 b) + (3 b) (– 4 c) + (– 4 c) (2a)]
= 4 a 2 + 9 b 2 + 16 c 2 + 2 (6 ab – 12 bc – 8 ca)
= 4 a 2 + 9 b 2 + 16 c 2 + 12 ab – 24 bc – 16 ca.
Example 2. Find the expansions of the following:


(i)  x + y 
3
7
2
5
2
2


(ii)  − 
2
3b
3a
5
2
2
2
2
1
2
2
2
4
5
5

 






Solution. (i)  x + y  =  x  + 2  x   y  +  y 
3
7
3
3
7
7
=
(ii)
2
4 2 20
25
x + xy + y 2 .
9
21
49
2 2
3a 2
3a
2
2 2
 3a
−  =   − 2     +   
 2   3b   3b 
 2
2
3b 
=
(iii)
2


(iii)  x − y − z  .
2
3
5
9 2
a
4
a −2 + 2 .
4
b 9b
1
 2
2
4 2
1
 x − y − z  =  2 x +  − 3
2
3
5
1
2
4

y  +  − z  
  5 
2
2
4
2






=  x  +  − y  +  − z 
2
3
5
 1
2

+ 2   x   − y  +  − y   − z  +  − z   x  
 2   3   3   5   5   2  
612
2
2
4
=
1 2 4 2 16 2
1
8
2
x + y + z + 2  − xy + yz − zx  4
9
25
15
5 
 3
=
1 2 4 2 16 2 2
16
4
x + y + z − xy + yz − zx .
4
9
25
3
15
5
Understanding ICSE mathematics – Ix
4
1
Example 3. Simplify the following:

(ii)  2x −
(i) (2x – 3y + 5) (2x – 3y – 5)
3
3
+ 1  2x + + 1 .


x
x
Solution. (i) Given expression = (2x – 3y + 5) (2x – 3y – 5).
Let 2 x – 3 y = a, then
given expression = (a + 5) (a – 5) = a 2 – 5 2 = a 2 – 25
= (2x – 3y) 2 – 25
= (2x) 2 – 2 (2x) (3y) + (3y) 2 – 25
= 4x 2 – 12xy + 9y 2 – 25.
3
x


3
x
(ii)Given expression= (2x + 1) −  (2x + 1) +  .
Let 2x + 1 = a, then
2
3 
3
 3

given expression =  a −   a +  = a2 −  
x
x
x
= (2x + 1) 2 –
9
x2
= (2x) 2 + 2 (2x) (1) + (1) 2 –
= 4x2 + 4x + 1 –
9
x2
9
x2
.
Example 4. Simplify the following:

(i)  2p −
q
q
− 3  2p + + 3


5
5
(ii) (3x – 2y) (3x + 2y) (9x2 + 4y2).

q


q

q
Solution. (i)Given expression =  2 p − − 3   2 p + + 3 
5
5
 
q

=  2 p −  5 + 3    2 p +  5 + 3  


q

= (2p)2 –  + 3 
5
= 4p2
2

  q  2
 q
2
+
2
(
3
)
+
(
3
)






–  5 
 5


= 4p2 –
q2 6 q
−
– 9.
25
5
(ii)Given expression = (3x – 2y) (3x + 2y)(9x2 + 4y2)
= ((3x – 2y) (3x + 2y)) (9x2 + 4y2)
= ((3x)2 – (2y)2) (9x2 + 4y2)
= (9x2 – 4y2) (9x2 + 4y2)
= (9x2)2 – (4y2)2
= 81x4 – 16y4.
Example 5. Find the expansions of the following:
(i) (2a + 3b)3
(ii) (3x – 4y) 3
(iii) (x + y – 1) 3.
Solution. (i) (2 a + 3 b) 3 = (2 a) 3 + (3 b) 3 + 3 (2 a) (3 b) (2 a + 3 b)
= 8 a 3 + 27 b 3 + 18 ab (2 a + 3 b)
= 8 a 3 + 27 b 3 + 36 a 2 b + 54 ab 2.
expansions
163
(ii)
(3x – 4y)3 = (3x) 3 – (4y) 3 – 3 (3 x) (4 y) (3 x – 4 y)
= 27 x 3 – 64 y 3 – 36xy (3x – 4y)
= 27 x 3 – 64 y 3 – 108 x 2 y + 144 xy 2.
(iii)
(x + y – 1)3 = (x + y − 1)3 [consider x + y as one term]
= (x + y)3 – (1)3 – 3(x + y) (1) (x + y − 1)
= (x + y)3 – 1 – 3(x + y) (x + y − 1)
= [x3 + y3 + 3xy (x + y)] – 1 – 3 (x + y)2 + 3(x + y)
= x3 + y3 + 3x2y + 3xy2 – 1 – 3 (x2 + 2xy + y2) + 3x + 3y
= x3 + y3 + 3x2y + 3xy2 – 3x2 – 6xy – 3y2 + 3x + 3y – 1.
Example 6. Find the product of the following (using standard results):
(i) (2x + 5y + 3) (2x + 5y + 4)
(ii) (3x – y – 4) (3x – y + 5).
Solution. (i) Given expression = (2x + 5y + 3) (2x + 5y + 4).
Let 2 x + 5 y = a, then
given expression = (a + 3) (a + 4)
= a 2 + (3 + 4) a + 3 × 4
[ (x + a) (x + b) = x 2 + (a + b) x + ab]
= a 2 + 7 a + 12
= (2 x + 5 y) 2 + 7 (2 x + 5 y) + 12
= (2 x) 2 + 2(2 x) (5 y) + (5 y)2 + 14 x + 35 y + 12
= 4 x 2 + 20 xy + 25 y 2 + 14 x + 35 y + 12
(ii)Given expression = (3 x – y – 4) (3 x – y + 5)
Let 3 x – y = a, then
given expression = (a – 4) (a + 5)
= a 2 – (4 – 5) a – 4 × 5
[ (x – a) (x + b) = x 2 – (a – b) x – ab]
= a 2 + a – 20
= (3 x – y) 2 + (3 x – y) – 20
= (3 x)2 – 2 (3 x) (y) + (y)2 + 3 x – y – 20
= 9 x 2 – 6 xy + y 2 + 3 x – y – 20.
Example 7. Simplify the following:
2 
∴
given expression = (3 x + 5 y) [(3 x) 2 – (3 x) (5 y) + (5 y) 2]
= (3 x) 3 + (5 y) 3
= 27 x 3 + 125 y 3.
(ii) We know that (a – b) (a 2 + ab + b 2) = a 3 – b 3,
2 
2
2 2
 


∴ given expression =  x −   x 2 + x. +   
x 
x
x 


 2
= (x) 3 –  
x
3
8
= x 3 – 3 .
x
614
4


(ii)  x −   x 2 + 2 + 2  .
x
x
Solution. (i) We know that (a + b) (a 2 – ab + b 2) = a 3 + b 3,
(i) (3x + 5y) (9x 2 – 15xy + 25y 2)
Understanding ICSE mathematics – Ix
Example 8. Simplify: (x + 3 y + 5z) (x 2 + 9 y 2 + 25 z 2 – 3 xy – 15yz – 5 zx).
Solution. We know that
(a + b + c) (a 2 + b 2 + c 2 – ab – bc – ca) = a 3 + b 3 + c 3 – 3abc.
∴Given expression = (x + 3 y + 5 z) [(x) 2 + (3 y) 2 + (5 z) 2 – (x) (3 y) – (3y)(5 z) – (5 z) (x)]
= (x) 3 + (3 y) 3 + (5 z) 3 – 3(x) (3 y) (5 z)
= x 3 + 27 y 3 + 125 z 3 – 45 xyz.
Example 9. Multiply (9x2 + 25y2 + 15xy + 12x – 20y + 16) by (3x – 5y – 4), using a suitable
identity.
Solution. We know that
(a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc.
∴ (3x – 5y – 4) (9x2 + 25y2 + 15xy + 12x – 20y + 16)
= (3x – 5y – 4) [(3x)2 + (– 5y)2 + (– 4)2 – (3x) (– 5y) – (– 5y) (– 4) – (– 4) (3x)]
= (3x)3 + (– 5y)3 + (– 4)3 – 3(3x) (– 5y) (– 4)
= 27x3 – 125y3 – 64 – 180xy
= 27x3 – 125y3 – 180xy – 64.
Example 10. Find the products of:
(i) (x + 2) (x + 3) (x + 5)
(ii) (x + 1) (x – 3) (x – 4).
Solution. We know that
(x + a) (x + b) (x + c) = x 3 + (a + b + c) x 2 + (ab + bc + ca) x + abc.
(i) Compare (x + 2) (x + 3) (x + 5) with (x + a) (x + b) (x + c).
Here a = 2, b = 3, c = 5.
∴Given product = x 3 + (2 + 3 + 5) x 2 + (2 × 3 + 3 × 5 + 5 × 2) x + 2 × 3 × 5
= x 3 + 10 x 2 + 31 x + 30.
(ii) Compare (x + 1) (x – 3) (x – 4) with (x + a) (x + b) (x + c).
Here a = 1, b = – 3 and c = – 4.
∴Given product = x 3 + (1 + (– 3) + (– 4)) x 2 + [1 . (– 3) + (– 3) (– 4) + (– 4) . 1] x
+ 1 . (– 3) (– 4)
= x 3 – 6x 2 + 5x + 12.
Example 11. Find the coefficient of x 2 and x in the product of (x – 5) (x + 3) (x + 7).
Solution. We know that
(x + a) (x + b) (x + c) = x 3 + (a + b + c) x 2 + (ab + bc + ca) x + abc.
Compare (x – 5) (x + 3) (x + 7) with (x + a) (x + b) (x + c).
Here a = – 5, b = 3 and c = 7.
∴ Coefficient of x 2 = a + b + c = (– 5) + 3 + 7 = 5 and
coefficient of x = ab + bc + ca = (– 5) × 3 + 3 × 7 + 7 × (– 5) = – 29.
Example 12. Find the coefficient of x2 in the expansion of (x2 + 2x + 3)2 + (x2 – 2x + 3)2.
Solution. Given expression = (x2 + 2x + 3)2 + (x2 – 2x + 3)2
= ((x2 + 3) + (2x))2 + ((x2 + 3) – (2x))2.
expansions
165
Let x2 + 3 = a and 2x = b, then
given expression = (a + b)2 + (a – b)2
= (a2 + 2ab + b2) + (a2 – 2ab + b2) = 2(a2 + b2)
= 2[(x2 + 3)2 + (2x)2]
= 2[(x2)2 + 2(x2) (3) + (3)2 + 4x2]
= 2(x4 + 6x2 + 9 + 4x2)
= 2x4 + 20x2 + 18.
∴ Coefficient of x2 = 20.
Example 13. By using (a + b)2 = a2 + 2ab + b2, find the value of (10·3)2.
Solution. (10·3)2 = (10 + 0·3)2
= (10)2 + 2 × 10 × 0·3 + (0·3)2
= 100 + 6 + ·09 = 106·09
Example 14. Using a suitable identity, find the value of (98)3.
Solution. We know that (a – b)3 = a3 – 3a2b + 3ab2 – b3.
(98)3 = (100 – 2)3
Now
= (100)3 – 3 × (100)2 × 2 + 3 × 100 × (2)2 – (2)3 (take a = 100 and b = 2)
= 1000000 – 60000 + 1200 – 8
= 1001200 – 60008 = 941192.
Example 15. If a + b + c = 0, prove that a3 + b3 + c3 = 3abc.
Solution. Given a + b + c = 0 ⇒ a + b = – c.
On cubing both sides, we get (a + b)3 = (– c)3
⇒ a3 + b3 + 3ab (a + b) = – c3 but a + b = – c
⇒ a3 + b3 + 3ab (– c) = – c3
⇒ a3 + b3 + c3 – 3abc = 0
⇒ a3 + b3 + c3 = 3abc.
Example 16. Without actually calculating the cubes, find the values of:
3
3
3
 
 
 
(ii)   +   −   .
4
3
12
1
(i) (– 23)3 + 153 + 83
1
7
Solution. (i) Let a = – 23, b = 15 and c = 8, then
a + b + c = – 23 + 15 + 8 = 0
∴a3 + b3 + c3 = 3abc
⇒(– 23)3
+
153
+
83
=
(See example 15)
3(– 23)
× 15 × 8 = – 8280.
7
1
(ii)Let a = 1 , b = and c = – , then
a + b + c =
∴
⇒
1 1
7
3+4−7
0
+ −
=
=
= 0.
4 3 12
12
12
a3 + b3 + c3 = 3abc
3
3
(See example 15)
3
3
3
 7
 7
 1
 1
 1
 1
  +   −   =   +   +  − 
4
3
12
4
3
12
616
12
3
4
1
1
7
3
7

  
= 3      −  = − .
4 3
12
48
Understanding ICSE mathematics – Ix
Example 17. If x = 2y + 6, then find the value of x3 – 8y3 – 36xy – 216.
Solution. Given x = 2y + 6 ⇒ x – 2y – 6 = 0
⇒ x + (– 2y) + (– 6) = 0
( if a + b + c = 0, then a3 + b3 + c3 = 3abc)
⇒ (x)3 + (– 2y)3 + (– 6)3 = 3(x) (– 2y) (– 6)
⇒ x3 – 8y3 – 216 = 36xy
⇒ x3 – 8y3 – 36xy – 216 = 0.
Example 18. If a = 1, b = – 2 and c = – 3, find the value of
a 3 + b 3 + c 3 − 3abc
ab + bc + ca − ( a 2 + b 2 + c 2 )
.
Solution. We know that (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc,
∴
a3 + b 3 + c 3 − 3 abc
ab + bc + ca − ( a2 + b 2 + c 2 )
=
( a + b + c)( a2 + b 2 + c 2 − ab − bc − ca)
− ( a2 + b 2 + c 2 − ab − bc − ca)
= – (a + b + c)
= – (1 – 2 – 3) = – (– 4) = 4.
Example 19. Using suitable identity, find the value of:
0·75 × 0·75 × 0·75 + 0·25 × 0·25 × 0·25
.
0·75 × 0·75 − 0·75 × 0·25 + 0·25 × 0·25
Solution. We know that a3 + b3 = (a + b) (a2 – ab + b2)
⇒
a3 + b 3
a2 − ab + b 2
= a + b.
Putting a = 0·75 and b = 0·25 in this result, we get
0·75 × 0·75 × 0·75 + 0·25 × 0·25 × 0·25
a3 + b 3
= 2
0·75 × 0·75 − 0·75 × 0·25 + 0·25 × 0·25
a − ab + b 2
= a + b = 0·75 + 0·25 = 1.
Example 20. If a + b + c = 0, find the value of
(b + c)2 (c + a)2 ( a + b)2
+
+
.
bc
ca
ab
Solution. Given a + b + c = 0 ⇒ a3 + b3 + c3 = 3 abc
…(i)
(See example 15)
Also a + b + c = 0 ⇒ b + c = – a, c + a = – b, a + b = – c.
(− a)2 (− b)2 (− c)2
2
2
2
∴ (b + c) + (c + a) + ( a + b) =
+
+
bc
ca
ab
bc
ca
ab
=
a2 b 2 c2
+
+
bc
ca
ab
=
a3 + b 3 + c3
3 abc
=
abc
abc
= 3.
Example 21. Simplify:
( a 2 − b 2 )3 + (b 2 − c 2 )3 + (c 2 − a 2 )3
( a − b ) 3 + ( b − c ) 3 + ( c − a )3
[Using (i)]
.
Solution. We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
Here, (a2 – b2) + (b2 – c2) + (c2 – a2) = 0,
∴ (a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3 = 3(a2 – b2) (b2 – c2) (c2 – a2).
Also, (a – b) + (b – c) + (c – a) = 0,
∴ (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a).
expansions
167
∴
( a2 − b 2 )3 + (b 2 − c 2 )3 + (c 2 − a2 )3
( a − b)3 + (b − c)3 + (c − a)3
=
3( a2 − b 2 )(b 2 − c 2 )(c 2 − a2 )
3( a − b)(b − c)(c − a)
( a − b)( a + b)(b − c)(b + c)(c − a)(c + a)
( a − b)(b − c)(c − a)
=
= (a + b) (b + c) (c + a).
Exercise 3.1
By using standard formulae, expand the following (1 to 9):
1. (i) (2x + 7y)2
2. (i)  3 x +  2x
3. (i)  3 x −  2x

1 2

1 2
2 2
1
(ii)  x + y  .
2
3
(ii) (3x2y + 5z)2.
3 2
1
(ii)  x − y  .
2
2
4. (i) (x + 3) (x + 5)
(ii) (x + 3) (x – 5)
(iii) (x – 7) (x + 9)
(iv) (x – 2y) (x – 3y).
2
5. (i) (x – 2y – z) (ii) (2x – 3y + 4z)2.
2
2
3
6. (i)  2x + − 1 
− 1 .
(ii)  x −

3
2x
7. (i) (x + 2)3
(ii) (2a + b)3.
8. (i)  3 x +  x
3y)3

x
13

9. (i) (5x –

2
3
(ii) (2x – 1)3.
3

1 
(ii)  2x − 3 y  .
Simplify the following (10 to 19):
10. (i) (a + b)2 + (a – b)2
2
1
1


11. (i)  a +  +  a −  a
a
1 2 
1 2

(ii)  a +  −  a −  .
a
a
12. (i) (3x – 1)2 – (3x – 2) (3x + 1)
(ii) (4x + 3y)2 – (4x – 3y)2 – 48xy.
13. (i) (7 p + 9 q) (7 p – 9 q)
3 
3

(ii)  2x −   2x +  .
x
x
14. (i) (2 x – y + 3) (2 x – y – 3)
(ii) (3 x + y – 5) (3 x – y – 5).
2
2
(ii) (a + b)2 – (a – b)2.
2



15. (i)  x + − 3   x − − 3  x
x
(ii) (5 – 2 x) (5 + 2 x) (25 + 4 x 2).
16. (i) (x + 2 y + 3) (x + 2 y + 7) (ii) (2 x + y + 5) (2 x + y – 9)
(iii) (x – 2y – 5) (x – 2y + 3) (iv) (3x – 4y – 2) (3x – 4y – 6).
17. (i) (2 p + 3 q) (4 p 2 – 6 pq + 9 q 2 )



(ii)  x +   x 2 − 1 + 2  .
x
x
18. (i) (3 p – 4 q) (9 p 2 + 12 pq + 16 q 2 )



(ii)  x −   x 2 + 3 + 2  .
x
x
1
3
1
9
19. (2x + 3 y + 4 z) (4 x 2 + 9 y 2 + 16 z 2 – 6 xy – 12 yz – 8 zx).
20. Find the product of the following:
(i) (x + 1) (x + 2) (x + 3) (ii) (x – 2) (x – 3) (x + 4).
21. Find the coefficient of x 2 and x in the product of (x – 3) (x + 7) (x – 4).
22. If a2 + 4a + x = (a + 2)2, find the value of x.
618
Understanding ICSE mathematics – Ix