1. Use suitable identities to find the following products:
(i) (x+4)(x +10)
Solution:
Using the identity, \((x+a)(x+b) = x^ 2+(a+b)x+ab\)[Here, a = 4 and b = 10]
\((x+4)(x+10) = x^2+(4+10)x+(4×10)\)
\(= x^2+14x+40\)
(ii) (x+8)(x –10)
Solution:
Using the identity, \((x+a)(x+b) = x ^2+(a+b)x+ab\)[Here, a = 8 and b = −10]
\((x+8)(x−10) = x^2+(8+(−10))x+(8×(−10))\)
\(= x^2+(8−10)x–80\)
\(= x2−2x−80\)
(iii) (3x+4)(3x–5)
Solution:
Using the identity, \((x+a)(x+b) = x^ 2+(a+b)x+ab\)[Here, x = 3x, a = 4 and b = −5]
\((3x+4)(3x−5) = (3x)^2+[4+(−5)]3x+4×(−5)\)
\(= 9x^2+3x(4–5)–20\)
\(= 9x^2–3x–20\)
(iv) \((y^2+{3\over2})(y^2-{3\over2})\)
Solution:
Using the identity, \((x+y)(x–y) = x^2–y ^2\)[Here, \(x = y^2\) and \(y = {3\over2}\)]
\((y2+3/2)(y2–3/2) = (y^2)^2–(3/2)^2\)
= y4–9/4
2. Evaluate the following products without multiplying directly:
(i) 103×107
Solution:
103×107= (100+3)×(100+7)
Using identity, [(x+a)(x+b) = x2+(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
Solution:
95×96 = (100-5)×(100-4)
Using identity, [(x-a)(x-b) = x2-(a+b)x+ab
Here, x = 100
a = -5
b = -4
We get, 95×96 = (100-5)×(100-4)
= (100)2+100(-5+(-4))+(-5×-4)
= 10000-900+20
= 9120
(iii) 104×96
Solution:
104×96 = (100+4)×(100–4)
Using identity, [(a+b)(a-b)= a2-b2]
Here, a = 100
b = 4
We get, 104×96 = (100+4)×(100–4)
= (100)2–(4)2
= 10000–16
= 9984
3. Factorize the following using appropriate identities:
(i) 9x2+6xy+y2
Solution:
9x2+6xy+y2 = (3x)2+(2×3x×y)+y2
Using identity, x2+2xy+y2 = (x+y)2
Here, x = 3x, y = y
9x2+6xy+y2 = (3x)2+(2×3x×y)+y2
= (3x+y)2
= (3x+y)(3x+y)
(ii) 4y2−4y+1
Solution:
4y2−4y+1 = (2y)2–(2×2y×1)+1
Using identity, x2 – 2xy + y2 = (x – y)2
Here, x = 2y
y = 1
4y2−4y+1 = (2y)2–(2×2y×1)+12
= (2y–1)2
= (2y–1)(2y–1)
(iii) x2–y2/100
Solution:
x2–y2/100 = x2–(y/10)2
Using identity, x2-y2 = (x-y)(x+y)
Here, x = x
y = y/10
x2–y2/100 = x2–(y/10)2
= (x–y/10)(x+y/10)
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) ((1/4)a-(1/2)b +1)2
Solution:
(i) (x+2y+4z)2
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = x
y = 2y
z = 4z
(x+2y+4z)2 = x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x2+4y2+16z2+4xy+16yz+8xz
(ii) (2x−y+z)2
Using identity, (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)+(2×z×2x)
= 4x2+y2+z2–4xy–2yz+4xz
(iii) (−2x+3y+2z)2
Solution:
Using identity, (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–8xz
(iv) (3a –7b–c)2
Solution:
Using identity (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
Solution:
Using 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
Solution:
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = (1/4)a
y = (-1/2)b
z = 1
5. Factorize:
(i) 4x2+9y2+16z2+12xy–24yz–16xz
(ii ) 2x2+y2+8z2–2√2xy+4√2yz–8xz
Solution:
(i) 4x2+9y2+16z2+12xy–24yz–16xz
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2
4x2+9y2+16z2+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) 2x2+y2+8z2–2√2xy+4√2yz–8xz
Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2
2x2+y2+8z2–2√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)
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 = x3+y3+3xy(x+y)
(2x+1)3= (2x)3+13+(3×2x×1)(2x+1)
= 8x3+1+6x(2x+1)
= 8x3+12x2+6x+1
(ii) (2a−3b)3
Using 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
Using 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)
(iv) (x−(2/3)y)3
Using identity, (x –y)3 = x3–y3–3xy(x–y)
7. Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solutions:
(i) (99)3
Solution:
We can write 99 as 100–1
Using identity, (x –y)3 = x3–y3–3xy(x–y)
(99)3 = (100–1)3
= (100)3–13–(3×100×1)(100–1)
= 1000000 –1–300(100 – 1)
= 1000000–1–30000+300
= 970299
(ii) (102)3
Solution:
We can write 102 as 100+2
Using identity,(x+y)3 = x3+y3+3xy(x+y)
(100+2)3 =(100)3+23+(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)3
Solution:
We can write 99 as 1000–2
Using identity,(x–y)3 = x3–y3–3xy(x–y)
(998)3 =(1000–2)3
=(1000)3–23–(3×1000×2)(1000–2)
= 1000000000–8–6000(1000– 2)
= 1000000000–8- 6000000+12000
= 994011992
8. 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)−(9/2) p2+(1/4)p
Solutions:
(i) 8a3+b3+12a2b+6ab2
Solution:
The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2
8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2
= (2a+b)3
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used.
(ii) 8a3–b3–12a2b+6ab2
Solution:
The expression, 8a3–b3−12a2b+6ab2 can be written as (2a)3–b3–3(2a)2b+3(2a)(b)2
8a3–b3−12a2b+6ab2 = (2a)3–b3–3(2a)2b+3(2a)(b)2
= (2a–b)3
= (2a–b)(2a–b)(2a–b)
Here, the identity,(x–y)3 = x3–y3–3xy(x–y) is used.
(iii) 27–125a3–135a+225a2
Solution:
The expression, 27–125a3–135a +225a2 can be written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2
27–125a3–135a+225a2 =
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 = x3–y3-3xy(x–y) is used.
(iv) 64a3–27b3–144a2b+108ab2
Solution:
The expression, 64a3–27b3–144a2b+108ab2can be written as (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2
64a3–27b3–144a2b+108ab2=
(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 = x3 – y3 – 3xy(x – y) is used.
(v) 27p3– (1/216)−(9/2) p2+(1/4)p
Solution:
The expression, 27p3–(1/216)−(9/2) p2+(1/4)p can be written as
(3p)3–(1/6)3−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6)
Using (x – y)3 = x3 – y3 – 3xy (x – y)
27p3–(1/216)−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6)
Taking x = 3p and y = 1/6
= (3p–1/6)3
= (3p–1/6)(3p–1/6)(3p–1/6)
9. Verify:
(i) x3+y3 = (x+y)(x2–xy+y2)
(ii) x3–y3 = (x–y)(x2+xy+y2)
Solutions:
(i) x3+y3 = (x+y)(x2–xy+y2)
We know that, (x+y)3 = x3+y3+3xy(x+y)
⇒ x3+y3 = (x+y)3–3xy(x+y)
⇒ x3+y3 = (x+y)[(x+y)2–3xy]
Taking (x+y) common ⇒ x3+y3 = (x+y)[(x2+y2+2xy)–3xy]
⇒ x3+y3 = (x+y)(x2+y2–xy)
(ii) x3–y3 = (x–y)(x2+xy+y2)
We know that,(x–y)3 = x3–y3–3xy(x–y)
⇒ x3−y3 = (x–y)3+3xy(x–y)
⇒ x3−y3 = (x–y)[(x–y)2+3xy]
Taking (x+y) common ⇒ x3−y3 = (x–y)[(x2+y2–2xy)+3xy]
⇒ x3+y3 = (x–y)(x2+y2+xy)
10. Factorize each of the following:
(i) 27y3+125z3
(ii) 64m3–343n3
Solutions:
(i) 27y3+125z3
The expression, 27y3+125z3 can be written as (3y)3+(5z)3
27y3+125z3 = (3y)3+(5z)3
We know that, 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
The expression, 64m3–343n3can be written as (4m)3–(7n)3
64m3–343n3 =
(4m)3–(7n)3
We know that, 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)
11. Factorise: \(27x^3+y^3+z^3–9xyz\)
Solution:
The expression \(27x^3+y^3+z^3–9xyz\) 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, \(x^3+y^3+z^3–3xyz = (x+y+z)(x^2+y^2+z^2–xy –yz–zx)\)
\(27x^3+y^3+z^3–9xyz = (3x)^3+y^3+z^3–3(3x)(y)(z)\)
\(= (3x+y+z)[(3x)2+y2+z2–3xy–yz–3xz]\)
\(= (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)\)
12. Verify that:
\(x^3+y^3+z^3–3xyz = {1\over 2} (x+y+z)[(x–y)^2+(y–z)^2+(z–x)^2]\)
Solution:
\(x^3+y^3+z^3−3xyz = (x+y+z)(x^2+y^2+z^2–xy–yz–xz)\)
\(x^3+y^3+z^3−3xyz = {1\over 2}(x+y+z)[2(x^2+y^2+z^2–xy–yz–xz)]\)
\(= {1\over 2}(x+y+z)(2x^2+2y^2+2z^2–2xy–2yz–2xz)\)
\(= {1\over 2}(x+y+z)[(x^2+y^2−2xy)+(y^2+z^2–2yz)+(x^2+z^2–2xz)]\)
\(= {1\over 2}(x+y+z)[(x–y)^2+(y–z)^2+(z–x)^2]\)
13. If x+y+z = 0, show that \(x^3+y^3+z^3 = 3xyz\).
Solution:
\(x^3+y^3+z^3 - 3xyz = {1 \over 2}(x +y+z)(x^2+y^2+z^2–xy–yz–xz)\)
ATQ
(x+y+z) = 0,
\(x^3+y^3+z^3 - 3xyz = (0)(x^2+y^2+z^2–xy–yz–xz)\)
\(x^3+y^3+z^3–3xyz = 0\)
\(x^3+y^3+z^3 = 3xyz\)
Hence Proved
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) \((28)^3+(−15)^3+(−13)^3\)
Solution:
\((28)^3+(−15)^3+(−13)^3\)
Let a = 28
b = −15
c = −13
We know that if x+y+z = 0, then \(x^3+y^3+z^3 = 3xyz\).
Here, \(x+y+z = 28–15–13 = 0\)
\((28)^3+(−15)^3+(−13)^3 = 3xyz\)
\(= 0+3(28)(−15)(−13)\)
\(= 16380\)
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)
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)
\(x^3+y^3+z^3–3xyz = 0\)
\(x^3+y^3+z^3 = 3xyz\)
Hence Proved
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) \((28)^3+(−15)^3+(−13)^3\)
Solution:
\((28)^3+(−15)^3+(−13)^3\)
Let a = 28
b = −15
c = −13
We know that if x+y+z = 0, then \(x^3+y^3+z^3 = 3xyz\).
Here, \(x+y+z = 28–15–13 = 0\)
\((28)^3+(−15)^3+(−13)^3 = 3xyz\)
\(= 0+3(28)(−15)(−13)\)
\(= 16380\)
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)
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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