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Q.1. Use suitable identities to find the following products:
(i) (x + 4)(x + 10)
(ii) (x + 8)(x – 10)
(iii) (3x + 4)(3x – 5)
(iv)
(v) (3 – 2x)(3 + 2x)
Solution:
(i) (x + 4)(x + 10):
Using the identity (x + a)(x + b) = x^{2} + (a + b)x + ab,
[Here, a = 4 and b = 10]
We get,
(x + 4)(x + 10) = x^{2} + (4 + 10)x + (4 x 10)
= x^{2} + 14x + 40
(ii) (x + 8)(x – 10)
Using the identity, (x+a)(x+b) = x^{2}+(a+b)x+ab
[Here, a = 8 and b = (–10)]
We get: (x + 8)(x – 10) = x^{2} + [8 + (–10)]x + [8 x (–10)]
= x^{2 }+ [810]x + [–80]
= x^{2} – 2x – 80
(iii) (3x + 4)(3x – 5)
Using the identity (x + a)(x + b) = x^{2} + (a + b)x + ab,
we get
(3x + 4)(3x – 5) = (3x)^{2} + [4 + (–5)]3x + [4 x (–5)]
= 9x^{2} + 3x(4–5)–20
= 9x^{2} – 3x – 20
(iv)
Using the identity (x + y)(x – y) = x^{2} – y^{2},
[Here, x = y^{2 }and y = 3/2]
we get:
(y^{2}+3/2)(y^{2}–3/2) = (y^{2})^{2}–(3/2)^{2}
= y^{4}–9/4
Q.2. Evaluate the following products without multiplying directly:
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i) 103×107= (100+3)×(100+7)
Using identity, [(x+a)(x+b) = x^{2}+(a+b)x+ab
Here, x = 100
a = 3
b = 7
We get, 103×107 = (100+3)×(100+7)
= (100)^{2}+(3+7)100+(3×7))
= 10000+1000+21
= 11021
(ii) 95×96 = (1005)×(1004)
Using identity, [(xa)(xb) = x^{2}(a+b)x+ab
Here, x = 100
a = 5
b = 4
We get, 95×96 = (1005)×(1004)
= (100)^{2}+100(5+(4))+(5×4)
= 10000900+20
= 9120
(iii) 104×96 = (100+4)×(100–4)
Using identity, [(a+b)(ab)= a2b2]
Here, a = 100
b = 4
We get, 104×96 = (100+4)×(100–4)
= (100)^{2}–(4)2
= 10000–16
= 9984
Q.3. Factorise the following using appropriate identities:
(i) 9x^{2} + 6xy + y^{2}
(ii) 4y^{2} – 4y + 1
(iii) x^{2}–y^{2}/100
Solution:
(i) 9x^{2}+6xy+y^{2} = (3x)^{2}+(2×3x×y)+y^{2}
Using identity, x^{2}+2xy+y^{2} = (x+y)^{2}
Here, x = 3x
y = y
9x^{2}+6xy+y^{2} = (3x)^{2}+(2×3x×y)+y^{2}
= (3x+y)^{2}
= (3x+y)(3x+y)
(ii) 4y^{2} – 4y + 1
4y^{2}−4y+1 = (2y)^{2}–(2×2y×1)+1
Using identity, x^{2} – 2xy + y^{2} = (x – y)^{2}
Here, x = 2y
y = 1
4y^{2}−4y+1 = (2y)^{2}–(2×2y×1)+12
= (2y–1)^{2}
= (2y–1)(2y–1)
(iii) x^{2}–y^{2}/100
x^{2}–y^{2}/100 = x^{2}–(y/10)^{2}
Using identity, x^{2}y^{2} = (xy)(x+y)
Here, x = x
y = y/10
x^{2}–y^{2}/100 = x^{2}–(y/10)^{2}
= (x–y/10)(x+y/10)
Q.4. 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)
Solution:
(i) (x + 2y + 4z)^{2}
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = x
y = 2y
z = 4z
(x+2y+4z)^{2} = x^{2}2+(2y)^{2}+(4z)^{2}+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x^{2}+4y^{2}+16z^{2}+4xy+16yz+8xz
(ii) (2x – y + z)^{2}
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = 2x
y = −y
z = z
(2x−y+z)^{2} = (2x)^{2}+(−y)^{2}+z^{2}+(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x^{2}+y^{2}+z^{2}–4xy–2yz+4xz
(iii) (–2x + 3y + 2z)^{2}
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+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)
= 4x^{2}+9y^{2}+4z^{2}–12xy+12yz–8xz
(iv) (3a – 7b – c)^{2}
Using identity (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+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)
= 9a^{2} + 49b^{2} + c^{2}– 42ab+14bc–6ca
(v) (–2x + 5y – 3z)^{2}
Using identity, (x+y+z)^{2}= x^{2}+y^{2}+z^{2}+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)
= 4x^{2}+25y^{2} +9z^{2}– 20xy–30yz+12zx
(vi)
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = (1/4)a
y = (1/2)b
z = 1
Q.5. Factorise:
(i) 4x^{2} + 9y^{2 }+ 16z^{2} + 12xy – 24yz – 16xz
(ii) 2x^{2} + y^{2 }+ 8z^{2 }– 2√2 xy + 4 2 yz – 8xz
Solution:
(i)4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}
4x^{2}+9y^{2}+16z^{2}+12xy–24yz–16xz = (2x)^{2}+(3y)^{2}+(−4z)^{2}+(2×2x×3y)+(2×3y×−4z) +(2×−4z×2x)
= (2x+3y–4z)^{2}
= (2x+3y–4z)(2x+3y–4z)
(ii) 2x^{2} + y^{2} + 8z^{2} – 2√2xy + 4√2 yz – 8xz
Using identity, (x +y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}
2x^{2}+y^{2}+8z^{2}–2√2xy+4√2yz–8xz
= (√2x)^{2}+(y)^{2}+(2√2z)^{2}+(2×√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z)_{2}
= (−√2x+y+2√2z)(−√2x+y+2√2z)
Q.6. 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 }
Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
(2x+1)3= (2x)^{3}+13+(3×2x×1)(2x+1)
= 8x^{3}+1+6x(2x+1)
= 8x^{3}+12x^{2}+6x+1
(ii) (2a – 3b)^{3}
Using identity,(x–y)^{3} = x^{3}–y^{3}–3xy(x–y)
(2a−3b)^{3} = (2a)^{3}−(3b)^{3}–(3×2a×3b)(2a–3b)
= 8a^{3}–27b^{3}–18ab(2a–3b)
= 8a^{3}–27b^{3}–36a^{2}b+54ab^{2}
(iii) ((3/2)x+1)^{3}
Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
((3/2)x+1)^{3}=((3/2)x)^{3}+13+(3×(3/2)x×1)((3/2)x +1)
(iv) (x−(2/3)y)^{3}
Using identity, (x –y)^{3} = x^{3}–y^{3}–3xy(x–y)
Q.7. Evaluate the following using suitable identities:
(i) (99)^{3 }
(ii) (102)^{3}
(iii) (998)^{3}
Solution:
(i) (99)^{3}
We have 99 = 100 – 1
Using identity, (x –y)^{3} = x^{3}–y^{3}–3xy(x–y)
∴ 99^{3} = (100 – 1)^{3}
= (100)^{3} – 1^{3} – (3x100x1)(100 – 1)
= 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 1000300 – 30001
= 970299
(ii) (102)^{3}
We can write 102 as 100+2
Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
(100+2)^{3} =(100)^{3}+2^{3}+(3×100×2)(100+2)
= 1000000 + 8 + 600[100 + 2]
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)^{3}
We can write 99 as 1000–2
Using identity,(x–y)^{3} = x^{3}–y^{3}–3xy(x–y)
(998)^{3} =(1000–2)^{3}
= 1000000000 – 8 – 6000[1000 – 2]
= 1000000000 – 8 – 6000000 – 12000
= 994011992
Q.8. Factorise each of the following:
(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}
(ii) 8a^{3} – b^{3} – 12a^{2}b + 6ab^{2 }
(iii) 27 – 125a^{3} – 135a + 225a^{2}
(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}
(v) 27p^{3}–(1/216)−(9/2)p^{2}+(1/4)p
Solution:
(i) 8a^{3} + b3 + 12a^{2}b + 6ab^{2}
The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2
8a^{3}+b^{3}+12a^{2}b+6ab^{2} = (2a)^{3}+b^{3}+3(2a)^{2}b+3(2a)(b)^{2}
= (2a+b)^{3}
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)^{3} = x^{3}+y^{3}+3xy(x+y) is used.
(ii)8a^{3 }– b^{3} – 12a^{2}b + 6ab^{2}
The expression, 8a^{3}–b^{3}−12a^{2}b+6ab^{2} can be written as (2a)^{3}–b^{3}–3(2a)^{2}b+3(2a)(b)^{2}
8a^{3}–b^{3}−12a^{2}b+6ab^{2} = (2a)^{3}–b^{3}–3(2a)^{2}b+3(2a)(b)^{2}
= (2a–b)^{3}
= (2a–b)(2a–b)(2a–b)
Here, the identity,(x–y)^{3} = x^{3}–y^{3}–3xy(x–y) is used.
(iii) 27 – 125a^{3} – 135a + 225a^{2}
The expression, 27–125a^{3}–135a +225a^{2} can be written as 3^{3}–(5a)^{3}–3(3)^{2}(5a)+3(3)(5a)^{2}
27–125a^{3}–135a+225a^{2} =
33–(5a)^{3}–3(3)^{2}(5a)+3(3)(5a)^{2}
= (3–5a)^{3}
= (3–5a)(3–5a)(3–5a)
Here, the identity, (x–y)^{3} = x^{3}–y^{3}3xy(x–y) is used.
(iv) 64a^{3} – 27b^{3} – 144 a^{2}b + 108 ab^{2}
The expression, 64a^{3}–27b^{3}–144a^{2}b+108ab^{2} can be written as
(4a)^{3}–(3b)^{3 } 3(4a)^{2}(3b)+3(4a)(3b)^{2}
64a^{3}–27b^{3 }– 144a^{2}b+108ab^{2}
= (4a)^{3}–(3b)^{3}–3(4a)^{2}(3b)+3(4a)(3b)^{2}
=(4a–3b)^{3}
=(4a–3b)(4a–3b)(4a–3b)
Here, the identity, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y) is used.
(v) 27p^{3}– (1/216)−(9/2) p^{2}+(1/4)p
The expression, 27p^{3}–(1/216)−(9/2) p^{2}+(1/4)p
can be written as (3p)^{3}–(1/6)3–3(3p)^{2}(1/6)+3(3p)(1/6)^{2}
27p^{3}–(1/216)−(9/2) p^{2}+(1/4)p =
(3p)^{3}–(1/6)^{3}–3(3p)^{2}(1/6)+3(3p)(1/6)^{2}
= (3p–16)^{3}
= (3p–16)(3p–16)(3p–16)
Q.9. Verify:
(i) x^{3} + y^{3} = (x + y)(x^{2 }– xy + y^{2})
(ii) x^{3 }– y^{3 }= (x – y)(x^{2} + xy + y^{2})
Solution:
(i) We know that, (x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
⇒ x^{3}+y^{3} = (x+y)^{3}–3xy(x+y)
⇒ x^{3}+y^{3} = (x+y)[(x+y)^{2}–3xy]
Taking (x+y) common ⇒ x^{3}+y^{3} = (x+y)[(x^{2}+y^{2}+2xy)–3xy]
⇒ x^{3}+y^{3} = (x+y)(x^{2}+y^{2}–xy)
(ii) We know that,(x–y)^{3} = x^{3}–y^{3}–3xy(x–y)
⇒ x^{3}−y^{3} = (x–y)^{3}+3xy(x–y)
⇒ x^{3}−y^{3} = (x–y)[(x–y)^{2}+3xy]
Taking (x+y) common ⇒ x^{3}−y^{3} = (x–y)[(x^{2}+y^{2}–2xy)+3xy]
⇒ x^{3}+y^{3} = (x–y)(x^{2}+y2+xy)
Q.10. Factorise each of the following:
(i) 27y^{3 }+ 125z^{3}
(ii) 64m^{3} – 343n^{3}
Solution:
(i) 27y^{3}+125z^{3}
The expression, 27y^{3}+125z^{3} can be written as (3y)^{3}+(5z)^{3}
27y^{3}+125z^{3} = (3y)^{3}+(5z)^{3}
We know that, x^{3}+y^{3} = (x+y)(x^{2}–xy+y2)
27y^{3}+125z^{3} = (3y)^{3}+(5z)^{3}
= (3y+5z)[(3y)^{2}–(3y)(5z)+(5z)^{2}]
= (3y+5z)(9y^{2}–15yz+25z^{2})
(ii) 64m^{3}–343n^{3}
The expression, 64m^{3}–343n^{3}can be written as (4m)^{3}–(7n)^{3}
64m^{3}–343n^{3} = (4m)^{3}–(7n)^{3}
We know that, x^{3}–y^{3} = (x–y)(x^{2}+xy+y^{2})
64m^{3}–343n^{3} = (4m)^{3}–(7n)^{3}
= (4m7n)[(4m)^{2}+(4m)(7n)+(7n)^{2}]
= (4m7n)(16m^{2}+28mn+49n^{2})
Q.11. Factorise 27x^{3 }+ y^{3 }+ z^{3} – 9xyz.
Solution:
The expression27x^{3}+y^{3}+z^{3}–9xyz can be written as (3x)^{3}+y^{3}+z^{3}–3(3x)(y)(z)
27x^{3}+y^{3}+z^{3}–9xyz = (3x)^{3}+y^{3}+z^{3}–3(3x)(y)(z)
We know that, x^{3}+y^{3}+z^{3}–3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}–xy –yz–zx)
27x^{3}+y^{3}+z^{3}–9xyz = (3x)^{3}+y^{3}+z^{3}–3(3x)(y)(z)
= (3x+y+z)[(3x)^{2}+y^{2}+z^{2}–3xy–yz–3xz]
= (3x+y+z)(9x^{2}+y^{2}+z^{2}–3xy–yz–3xz)
Q.12. Verify that x3 + y3 + z3 – 3xyz = (1/2) (x + y + z)[(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]
Solution:
We know that,
x^{3}+y^{3}+z^{3}−3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}–xy–yz–xz)
⇒ x^{3}+y^{3}+z^{3}–3xyz = (1/2)(x+y+z)[2(x^{2}+y^{2}+z^{2}–xy–yz–xz)]
= (1/2)(x+y+z)(2x^{2}+2y^{2}+2z^{2}–2xy–2yz–2xz)
= (1/2)(x+y+z)[(x^{2}+y^{2}−2xy)+(y^{2}+z^{2}–2yz)+(x^{2}+z^{2}–2xz)]
= (1/2)(x+y+z)[(x–y)^{2}+(y–z)^{2}+(z–x)^{2}]
Q.13. If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz.
Solution:
We know that,
x^{3}+y^{3}+z^{3}3xyz = (x +y+z)(x^{2}+y^{2}+z^{2}–xy–yz–xz)
Now, according to the question, let (x+y+z) = 0,
then, x^{3}+y^{3}+z^{3} 3xyz = (0)(x^{2}+y^{2}+z^{2}–xy–yz–xz)
⇒ x^{3}+y^{3}+z^{3}–3xyz = 0
⇒x^{3}+y^{3}+z^{3} = 3xyz
Hence Proved
Q.14. Without actually calculating the cubes, find the value of each of the following: (i) (–12)^{3} + (7)^{3} + (5)^{3}
(ii) (28)^{3} + (–15)^{3} + (–13)^{3}
Solution:
(i) (–12)^{3} + (7)^{3 }+ (5)^{3}
Let a = −12
b = 7
c = 5
We know that if x+y+z = 0, then x^{3}+y^{3}+z^{3}=3xyz.
Here, −12+7+5=0
(−12)^{3}+(7)^{3}+(5)^{3} = 3xyz
= 3×12×7×5
= 1260
(ii) (28)^{3} + (–15)^{3} + (–13)^{3 }
Let a = 28
b = −15
c = −13
We know that if x+y+z = 0, then x^{3}+y^{3}+z^{3} = 3xyz.
Here, x+y+z = 28–15–13 = 0
(28)^{3}+(−15)^{3}+(−13)^{3} = 3xyz
= 0+3(28)(−15)(−13)
= 16380
Q.15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a^{2}–35a+12
(ii) Area : 35y^{2}+13y–12
Solution:
(i) Area : 25a^{2}–35a+12
Using the splitting the middle term method,
We have to find a number whose sum = 35 and product =25×12=300
We get 15 and 20 as the numbers [15+20=35 and 15×20=300]
25a^{2}–35a+12 = 25a^{2}–15a−20a+12
= 5a(5a–3)–4(5a–3)
= (5a–4)(5a–3)
Possible expression for length = 5a–4
Possible expression for breadth = 5a –3
(ii) Area : 35y^{2}+13y–12
Using the splitting the middle term method,
We have to find a number whose sum = 13 and product = 35×12 = 420
We get 15 and 28 as the numbers [15+28 = 13 and 15×28=420]
35y^{2}+13y–12 = 35y^{2}–15y+28y–12
= 5y(7y–3)+4(7y–3)
= (5y+4)(7y–3)
Possible expression for length = (5y+4)
Possible expression for breadth = (7y–3)
Ques 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x^{2}–12x
(ii) Volume : 12ky^{2}+8ky–20k
Solution:
(i) Volume : 3x^{2}–12x
3x^{2}–12x can be written as 3x(x–4) by taking 3x out of both the terms.
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x–4)
(ii) Volume: 12ky^{2}+8ky–20k
12ky^{2}+8ky–20k can be written as 4k(3y^{2}+2y–5) by taking 4k out of both the terms.
12ky^{2}+8ky–20k = 4k(3y^{2}+2y–5)
[Here, 3y^{2}+2y–5 can be written as 3y^{2}+5y–3y–5 using splitting the middle term method.]
= 4k(3y^{2}+5y–3y–5)
= 4k[y(3y+5)–1(3y+5)]
= 4k(3y+5)(y–1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y 1)
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