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RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

Question 1.
Find the following products:
(i) (3x + 2y) (9X² – 6xy + Ay²)
(ii) (4x – 5y) (16x² + 20xy + 25y²)
(iii) (7p⁴ + q) (49p⁸ – 7p⁴q + q²)

Solution:

Question 2.
If x = 3 and y = -1, find the values of each of the following using in identity:

Solution:

Question 3.
If a + b = 10 and ab = 16, find the value of a² – ab + b² and a² + ab + b².
Solution:
a + b = 10, ab = 16 Squaring,
(a + b)² = (10)²
⇒ a² + b² + lab = 100
⇒ a² + b² + 2 x 16 = 100
⇒  a² + b² + 32 = 100
∴ a² + b² = 100 – 32 = 68
Now, a² – ab + b² = a² + b² – ab = 68 – 16 = 52
and a² + ab + b² = a² + b² + ab = 68 + 16 = 84

Question 4.
If a + b = 8 and ab = 6, find the value of a³ + b³.
Solution:
a + b = 8, ab = 6
Cubing both sides,
(a + b)³ = (8)3
⇒ a³ + b³ + 3 ab{a + b) = 512
⇒  a³ + b³ + 3 x 6 x 8 = 512
⇒  a³ + b³ + 144 = 512
⇒  a³ + b³ = 512 – 144 = 368
∴ a³ + b³ = 368

Question 5.
If a – b = 6 and ab = 20, find the value of a³-b³.
Solution:
a – b = 6, ab = 20
Cubing both sides,
(a – b)3 = (6)³
⇒  a³ – b³ – 3ab(a – b) = 216
⇒  a³ – b³ – 3 x 20 x 6 = 216
⇒  a³ – b³ – 360 = 216
⇒  a³ -b³ = 216 + 360 = 576
∴ a³ – b³ = 576

Question 6.
If x = -2 and y = 1, by using an identity find the value of the following:

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.3

Other Exercises

• RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.1
• RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.2
• RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.3
• RD Sharma Class 8 Solutions Chapter 2 Powers MCQS

Question 1.
Express the following numbers in standard form :
(i) 6020000000000000
(ii) 0.00000000000942
(iii) 0.00000000085
(iv) 846 X 10⁷
(v) 3759 x 10^-4
(vi) 0.00072984
(vii) 0.000437 x 10⁴ 
(Viii) 4 + 100000
Solution:

Question 2.
Write the following numbers in the usual form :
(i) 4.83 x 10⁷
(ii) 3.02 x 10^-6
(iii) 4.5 x 10⁴
(iv) 3 x 10^-8
(v) 1.0001 x 10⁹
(vi) 5.8 x 10²
(vii) 3.61492 x 10⁶
(viii) 3.25 x 10^-7
Solution:

Hope given RD Sharma Class 8 Solutions Chapter 2 Powers Ex 2.3 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

Question 1.
Find the cube of each of the following binomial expressions:

Solution:

Question 2.
If a + b = 10 and ab = 21, find the value of a³ + b³.
Solution:
a + b = 10, ab = 21
Cubing both sides,
(a + b)³ = (10)³
⇒ a³ + 6³ + 3ab (a + b) = 1000
⇒  a³ + b³ + 3 x 21 x 10 = 1000
⇒  a³ + b³ + 630 = 1000
⇒  a³ + b³ = 1000 – 630 = 370
∴ a³ + b³ = 370

Question 3.
If a – b = 4 and ab = 21, find the value of a³-b³.
Solution:
a – b = 4, ab= 21
Cubing both sides,
⇒ (a – A)³ = (4)³
⇒ a³ – b³ – 3ab (a – b) = 64
⇒ a³-i³-3×21 x4 = 64
⇒  a³ – 6³ – 252 = 64
⇒  a³ – 6³ = 64 + 252 =316
∴ a³ – b³ = 316

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If 2x + 3y = 13 and xy = 6, find the value of 8x³ + 21y³.
Solution:
2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y)³ = (13)³
⇒ (2x)³ + (3y)³ + 3 x 2x x 3X2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18xy(2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18 x 6 x 13 = 2197
⇒ 8X³ + 27y³ + 1404 = 2197
⇒  8x³ + 27y³ = 2197 – 1404 = 793
∴ 8x³ + 27y³ = 793

Question 10.
If 3x – 2y= 11 and xy = 12, find the value of 27x³ – 8y³.
Solution:
3x – 2y = 11 and xy = 12 Cubing both sides,
(3x – 2y)³ = (11)³
⇒  (3x)³ – (2y)³ – 3 x 3x x 2y(3x – 2y) =1331
⇒  27x³ – 8y³ – 18xy(3x -2y) =1331
⇒   27x³ – 8y³ – 18 x 12 x 11 = 1331
⇒  27x³ – 8y³ – 2376 = 1331
⇒  27X³ – 8y³ = 1331 + 2376 = 3707
∴ 2x³ – 8y³ = 3707

Question 11.
Evaluate each of the following:
(i)  (103)³
(ii) (98)³
(iii) (9.9)³
(iv) (10.4)³
(v) (598)³
(vi) (99)³
Solution:
We know that (a + bf = a³ + b³ + 3ab(a + b) and (a – b)³= a³ – b³ – 3 ab(a – b)
Therefore,
(i)  (103)³ = (100 + 3)³
= (100)³ + (3)³ + 3 x 100 x 3(100 + 3)    {∵ (a + b)³ = a³ + b³ + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98)³ = (100 – 2)³
= (100)³ – (2)³ – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9)³ = (10 – 0.1)³
= (10)³ – (0.1)³ – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4)³ = (10 + 0.4)³
= (10)³ + (0.4)³ + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)³ = (600 – 2)³
= (600)³ – (2)³ – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99)³ = (100 – 1)³
= (100)³ – (1)³ – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299

Question 12.
Evaluate each of the following:
(i)  111³ – 89³
(ii) 46³ + 34³
(iii) 104³ + 96³
(iv) 93³ – 107³
Solution:
We know that a³ + b³ = (a + bf – 3ab(a + b) and a³ – b³ = (a – bf + 3 ab(a – b)
(i) 111³ – 89³
= (111 – 89)³ + 3 x ill x 89(111 – 89)
= (22)³ + 3 x 111 x 89 x 22
= 10648 + 652014 = 662662
(OR)
(a + b)³ – (a – b)³ = 2(b³ + 3a²b)
= 111³ – 89³ = (100 + 11)³ – (100 – 11)³
= 2(11³ + 3 x 100² x 11]
= 2(1331 + 330000]
= 331331 x 2 = 662662
(a + b)³ + (a- b)³ = 2(b³ + 3ab²)
(ii) 46³ + 34³ = (40 + 6)³ + (40 – 6)³
= 2[(40)³ + 3 x 40 x 6²]
= 2[64000 + 3 x 40 x 36]
= 2[64000 + 4320]
= 2 x 68320 = 136640
(iii) 104³ + 96³ = (100 + 4)³ + (100 – 96)³
= 2 [a³ + 3 ab²]
= 2[(100)³ + 3 x 100 x (4)²]
= 2[ 1000000 + 300 x 16]
= 2[ 1000000 + 4800]
= 1004800 x 2 = 2009600
(iv) 93³ – 107³ = -[(107)³ – (93)³]
= -[(100 + If – (100 – 7)³]
= -2[b³ + 3a²b)]
= -2[(7)³ + 3(100)² x 7]
= -2(343 + 3 x 10000 x 7]
= -2[343 + 210000]
= -2[210343] = -420686

Question 13.

Solution:

Question 14.
Find the value of 27X³ + 8y³ if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = \(\frac { 14 }{ 9 }\)
Solution:

Question 15.
Find the value of 64x³ – 125z³, if 4x – 5z = 16 and xz = 12.
Solution:
4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z)³ = (16)³
⇒ (4x)³ – (5y)³ – 3 x 4x x 5z(4x – 5z) = 4096
⇒ 64x³ – 125z³ – 3 x 4 x 5 x xz(4x – 5z) = 4096
⇒  64x³ – 125z³ – 60 x 12 x 16 = 4096
⇒ 64x³ – 125z³ – 11520 = 4096
⇒  64x³ – 125z³ = 4096 + 11520 = 15616

Question 16.

Solution:

Question 17.
Simplify each of the following:

Solution:

Question 18.

Solution:

Question 19.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.