- Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y2+ 3/2 ) (y2– 3/2 )
(v) (3 – 2x) (3 + 2x)
Solution:
(i) (x + 4) (x + 10)
Using identity (x + a)(x + b) = x2 + (a + b)x + ab
(x + 4)(x + 10) = x2 + (4 + 10)x + 4×10
= x2 + 14x + 40
(ii) (x + 8) (x – 10)
Using identity (x + a)(x + b) = x2 + (a + b)x + ab
(x + 8) (x – 10) = x2 + (8 – 10)x + 8x(-10)
= x2 – 2x – 80x
(iii) (3x + 4) (3x – 5)
Using identity (x + a)(x + b) = x2 + (a + b)x + ab
(3x + 4) (3x – 5) = (3x)2 + (4 – 5)3x + 4x(-5)
= 9x2 – 3x – 20
(iv) (y2+ 3/2 ) (y2– 3/2 )
Using identity (x + a)(x – a) = x2 – a2
(y2+ 3/2 ) (y2– 3/2 ) = y2 – (3/2)2
= y2 – 9/4
(v) (3 – 2x) (3 + 2x)
Using identity (x + a)(x – a) = x2 – a2
(3 – 2x) (3 + 2x)= 32 – (2x)2 = 92 – 4x2
- Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) 103 × 107 = (100 + 3)(100 + 7)
Using identity (x + a)(x + b) = x2 + (a + b)x + ab
(100 + 3)(100 + 7) = 1002 + (3 + 7)100 + 3×7
= 10000 + 10(100) + 21
= 10000 + 1000 + 21
= 11021
(ii) 95 × 96
= (100 – 5)(100 – 4)
Using identity (x + a)(x + b) = x2 + (a + b)x + ab
(100 – 5)(100 – 4) = 1002 + (-5 – 4)*100 + (-4)(-5)
= 10000 – 900 + 20
= 9120
(iii) 104 × 96
=(100 + 4)(100 – 4)
Using the identity (x + a)(x – a) = x2 – a2
(100 + 4)(100 – 4) = 1002 – 42 = 10000 – 16 = 9984
- Factorise the following using appropriate identities:
(i) 9x2+ 6xy + y2
(ii) 4y2– 4y + 1
(iii) x2 – 1/100 .y2
Solution:
(i) 9x2+ 6xy + y2
=(3x)2 + 2(3x)(y) + y2
Using the identity a2 + 2ab + b2 = (a + b)2
= (3x + y)2
= (3x + y)(3x + y)
(ii) 4y2– 4y + 1
= (2y)2 – 2(2y)(1) + 12
Using the identity a2 + 2ab + b2 = (a + b)2
= (2y – 1)2
= (2y – 1)(2y – 1)
(iii) x2 – 1/100 .y2
= x2 – 2.x.1/10y + (1/10y)2
Using the identity a2 + 2ab + b2 = (a + b)2
= (x – 1/10y)2
=(x – 1/10y) (x – 1/10y)
- Expand each of the following using suitable identities:
(i) (x + 2y + 4z)2
(ii) (2x – y + z)2
(iii) (-2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v) (-2x + 5y – 3z)2
(vi) [1/4 a – 1/2 b + 1]2
Solution:
(i) (x + 2y + 4z)2
Identity V : (x + y + z)2= x2+ y2+ z2+ 2xy + 2yz + 2zx
here, x = x ; y = 2y ; z = 4z
= x2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x2 + 4y2 + 4z2 + 4xy + 16yz + 8xz
(ii) (2x – y + z)2
Identity V : (x + y + z)2= x2+ y2+ z2+ 2xy + 2yz + 2zx
here, x = 2x ; y = -y ; z = z
(2x – y + z)2= (2x)2+ (-y)2+ z2+ 2(2x)(-y) + 2(-y)z + 2z(2x)
= 4x2 – y2 + z2 – 4xy – 2yz + 4xz
(iii) (-2x + 3y + 2z)2
Identity V : (x + y + z)2= x2+ y2+ z2+ 2xy + 2yz + 2zx
Here, x = -2x ; y = 3y ; z = 2z
(-2x + 3y + 2z)2= (-2x)2+ (3y)2+ (2z)2+ 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx
(iv) (3a – 7b – c)2
Identity V : (x + y + z)2= x2+ y2+ z2+ 2xy + 2yz + 2zx
Here x = 3a ; y = -7b ; z = -c
(3a – 7b – c)2= (3a)2+ (-7b)2+ (-c)2+ 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca
(v) (-2x + 5y – 3z)2
Identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here x = -2x ; y = 5y ; z = -3z
(-2x + 5y – 3z)2 = (-2x)2 + (5y)2 + (-3z)2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx
(vi) [1/4 a – 1/2 b + 1]2
Identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here x =1/4 a ; y = – 1/2 b ; z = 1
[1/4 a – 1/2 b + 1]2 = (1/4 a)2 + (– 1/2 b)2 + 12 + 2(1/4 a)( – 1/2 b) + 2(– 1/2 b)(1) + 2(1)( 1/4 a)
= 1/16 a2 + 1/4b2 + 1 – 1/4 ab – b + 1/2 a
- Factorise:
(i) 4x2+ 9y2+ 16z2+ 12xy – 24yz – 16xz
(ii) 2x2 + y2+ 8z2– 2√2 xy + 4√2 yz – 8xz
Solution:
(i) 4x2+ 9y2+ 16z2+ 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (4z)2 + 2(2x)(3y) + 2(3y)4z) + 2(4z)(2x)
Identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= (2x + 3y + z)2
(ii) 2x2 + y2+ 8z2– 2√2 xy + 4√2 yz – 8xz
= (√2x)2 + y2 + (2√2z)2 + 2(√2x)y + 2(2√2z)y + 2(2√2z)(√2x)
Identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= (√2x + y + 2√2z)2
- Write the following cubes in expanded form:
(i) (2x + 1)3
(ii) (2a – 3b)3
(iii) [3/2 x + 1]3
(iv) [ x – 2/3 y]3
Solution:
(i) (2x + 1)3
Identity : (x + y)3 = x3 + y3 + 3xy(x + y)
(2x + 1)3 = (2x)3 + 13 + 3(2x)(1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 1+ 12x2 + 6x
= 8x3 + 12x2 + 6x + 1
(ii) (2a – 3b)3
Identity : (x – y)3 = x3 – y3 – 3xy(x – y)
(2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b)
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2
(iii) [3/2 x + 1]3
Identity : (x + y)3 = x3 + y3 + 3xy(x + y)
[3/2 x + 1]3 = (3/2 x)3+ 13 + 3(3/2 x)(1)(3/2 x + 1)
= 27/8 x3 + 1 + 9/2 x(3/2 x + 1)
= 27/8 x3 + 1 + 27/4 x + 9/2 x
(iv) [ x – 2/3 y]3
Identity : (x – y)3 = x3 – y3 – 3xy(x - y)
[ x – 2/3 y]3 = x3 – (2/3y)3 – 3(x)(2/3y)(x – 2/3y)
= x3 – 8/27 y3 – 2x2y + 4/3xy2
- Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) (99)3 = (100 – 1)3
Identity : (x – y)3 = x3 – y3 – 3xy(x – y)
(100 – 1)3 = 1003 – 13 – 3(100)(1)(100 – 1)
= 10000 – 1 – 300(100 – 1)
= 10000 – 1 – 30000 – 300
= 970299
(ii) (102)3 = (100 + 2)3
Identity : (x + y)3 = x3 + y3 + 3xy(x + y)
(100 + 2)3 = 1003 + 23 + 3*100*2(100 + 2)
= 10000 + 8 + 600(100 + 2)
= 10000 + 8 + 60000 + 1200
= 1061208
(iii) (998)3
Identity : (x – y)3 = x3 – y3 – 3xy(x – y)
(1000 – 2)3 = 10003 – 23 – 3*1000*2(1000 – 2)
= 1000000 – 8 – 6000(1000 – 2)
=1000000 – 8 – 6000000 + 12000
= 994011992
- Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2
(v) 27p3 – 1/216 – 9/2p2 + 1/4 p
Solution:
(i) 8a3 + b3 + 12a2b + 6ab2
Identity : (x + y)3 = x3 + y3 + 3xy(x + y)
= (2a)3 + b3 + 3(2a)(b) (2a + b)
= (2a + b)3
=(2a + b) (2a + b) (2a + b)
(ii) 8a3 – b3 – 12a2b + 6ab2
Identity : (x – y)3 = x3 – y3 – 3xy(x – y)
= (2a)3 – (b)3 – 3(2a)(b)(ab – b)
= (2a – b)3
=(2a + b) (2a + b) (2a + b)
(iii) 27 – 125a3 – 135a + 225a2
Identity : (x – y)3 = x3 – y3 – 3xy(x – y)
=(3)3– (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
(iv) 64a3 – 27b3 – 144a2b + 108ab2
Identity : (x – y)3 = x3 – y3 – 3xy(x – y)
=(4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
=(4a – 3b)3
=(4a – 3b) (4a – 3b) (4a – 3b)
(v) 27p3 – 1/216 – 9/2p2 + 1/4 p
Identity : (x – y)3 = x3 – y3 – 3xy(x – y)
=(3p)3 – (1/6)3 – 3(3p)(1/6)(3p – 1/6)
= (3p – 1/6)3
=(3p – 1/6) (3p – 1/6) (3p – 1/6)
- Verfiy:
(i) x3 + y3 = (x + y)(x2 – xy + y2)
(ii) x3 – y3 = (x – y)(x2 + xy + y2)
Solution:
(i) x3 + y3 = (x + y)(x2 – xy + y2)
R.H.S = (x + y)(x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
= L.H.S
(ii) x3 – y3 = (x – y)(x2 + xy + y2)
R.H.S. = (x – y)(x2 + xy + y2)
= x(x2 + xy + y2) – y(x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3
= L.H.S
- Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
Solution:
(i) 27y3 + 125z3
Using the identity: x3 + y3 = (x + y)(x2 – xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2)
(ii) 64m3 – 343n3
sing the identity: x3 – y3 = (x – y)(x2 + xy + y2)
64m3 – 343n3 = (4m)3– (7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2)
- Factorise 27x3 + y3 + z3 – 9xyz
Solution:
27x3 + y3 + z3 – 9xyz = (3x)3 + y3 + z3 – 3(3x)(y)(z)
Using identity: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
a = 3x ; b = y ; c = z
=(3x + y + z)[(3x)2 + (y)2 + (z)2 – (3x)(y) – (y)(z) – (z)(3x)]
= (3x + y + z)( 9x2 + y2 + z2 – 3xy – yz – 3xz)
- Verify that: x3 + y3 + z3 – 3xyz = 1/2(x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]
Solution:
R.H.S. = 1/2(x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]
= 1/2(x + y + z)[x2 – 2xy + y2 + y2 – 2yz + z2 + z2 – 2zx + x2]
= 1/2(x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)
= 1/2 (x + y + z)2(x2 + y2 + z2 – xy – yz – zx)
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
=x(x2 + y2 + z2 – xy – yz – zx) + y(x2 + y2 + z2 – xy – yz – zx) + z(x2 + y2 + z2 – xy – yz – zx)
= x3 + xy2 + xz2 – x2 y – xyz – zx2 + x2y + y3 + z2y – xy2 – y2z – xyz + x2z + zy2 + z3 – xyz – yz2 – xz2
= x3 + y3 + z3 – 3xyz
- If x + y + z = 0 show that x3 + y3 + z3 = 3xyz
Solution:
We know the identity x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) ………….(*)
It is given that x + y + z = 0 ……………(1)
Substitute eq(1) in (*)
x3 + y3 + z3 – 3xyz = (0)(x2 + y2 + z2 – xy – yz – zx)
x3 + y3 + z3 – 3xyz = 0
Therefore, x3 + y3 + z3 = 3xyz
- Without actually calculating the cubes, find the value of each of the following:
(i) (-12)3 + 73 + 53
(ii)(28)3 + (-15)3 + (-13)3
Solution:
(i) (-12)3 + 73 + 53
We know that x3 + y3 + z3 = 3xyz if x + y + z = 0
Here, -12 + 7 + 5 = 0
(-12)3 + 73 + 53 = 3(-12)(7)(5) = – 1260
(ii)(28)3 + (-15)3 + (-13)3
We know that x3 + y3 + z3 = 3xyz if x + y + z = 0
Here, 28 – 15 – 13 = 0
(28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) = 16380
- Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) area : 25a2 – 35a + 12
(ii) area : 35y2 + 13y – 12
Solution:
(i) area = 25a2 – 35a + 12
= 25a2 – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
=(5a – 4)(5a – 3)
So, one possible answer is length = (5a – 4), breadth = (5a – 3)
(ii) area = 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
=(5y + 4)(7y – 3)
So, (5y + 4) can be taken as breadth and (7y – 3) as length.
- What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: 3x2 – 12x
(ii) Volume 12ky2 + 8ky – 20 k
Solution:
(i) Volume: 3x2 – 12x
abc = 3x2 – 12x = 3x(x – 4)
Therefore, 3, x and (x – 4) are the three factors so they can be three dimensions
(ii) Volume 12ky2 + 8ky – 20 k
abc = 12ky2 + 8ky – 20k
= 4k(3y2 + 2y – 5)
= 4k (3y2 – 3y + 5y – 5)
= 4k (3y(y – 1)+5( y – 1))
= 4k(3y + 5)(y – 1)
Therefore, 4k , (y – 1), (3y + 5) are the three factors of three dimensions.
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