RD Sharma Class 9 Solutions Chapter 4 Exercise 4.5 (Updated for 2021-22)

RD Sharma Class 9 Solutions Chapter 4 Exercise 4.5

RD Sharma Class 9 Solutions Chapter 4 Exercise 4.5: Download the Free PDF of RD Sharma Class 9 Solutions Chapter 4 Exercise 4.5 and get good marks in your Class 9 Maths exams. You can always rely on this book as all the solutions are designed by subject matter experts in an easy language with step by step explanation. To know more about the RD Sharma Solutions Class 9 Maths, read the whole blog. 

Download RD Sharma Class 9 Solutions Chapter 4 Exercise 4.5 PDF

RD Sharma Class 9 Solutions Chapter 4 Exercise 4.5

 


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Question 1.
Find the following products:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
(ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx)
(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
(iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz)
Solution:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
= (3x + 2y + 2z) [(3x)2 + (2y)2 + (2z)2 – 3x x 2y + 2y x 2z + 2z x 3x]
= (3x)3 + (2y)3  + (2z)3 – 3 x 3x x 2y x 2z
= 27x3 + 8y3 + 8Z3 – 36xyz
(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)
= (4x -3y + 2z) [(4x)2 + (-3y)2 + (2z)2 – 4x x (-3y + (3y) x (2z) – (2z x 4x)]
= (4x)3 + (-3y)3 + (2z)3 – 3 x 4x x (-3y) x (2z)
= 64x3 – 27y3 + 8z3 + 72xyz
(iii) (2a -3b- 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
= (2a -3b- 2c) [(2a)2 + (3b)2 + (2c)2 – 2a x (-3b) – (-3b) x (-2c) – (-2c) x 2a]
= (2a)3 + (3b)3 + (-2c)3 -3x 2a x (-3 b) (-2c)
= 8a3 – 21b3 -8c3 – 36abc
(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)
= [3x + (-4y) + 5z] [(3x)2 + (-4y)2 + (5z)– 3x x (-4y) -(-4y) (5z) – 5z x 3x]
= (3x)3 + (-4y)3 + (5z)3 – 3 x 3x x (-4y) (5z)
= 27x3 – 64y3 + 125z3 + 180xyz

Question 2.
Evaluate:
Solution Of Rd Sharma Class 9 Chapter 4 Algebraic Identities
Solution:
RD Sharma Class 9 Chapter 4 Algebraic Identities
RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

Question 3.
If x + y + z = 8 and xy + yz+ zx = 20, find the value of x3 + y3 + z3 – 3xyz.
Solution:
We know that
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 -xy -yz – zx)
Now, x + y + z = 8
Squaring, we get
(x + y + z)2 = (8)2
x2 + y2 + z2 + 2(xy + yz + zx) = 64
⇒ x2 + y2 + z2 + 2 x 20 = 64
⇒  x2 + y2 + z2 + 40 = 64
⇒  x2 + y2 + z2 = 64 – 40 = 24
Now,
x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)]
= 8(24 – 20) = 8 x 4 = 32

Question 4.
If a +b + c = 9 and ab + bc + ca = 26, find the value of a3 + b3 + c3 – 3abc.
Solution:
a + b + c = 9, ab + be + ca = 26
Squaring, we get
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒  a2 + b2 + c2 + 2 x 26 = 81
⇒ a2 + b2 + c2 + 52 = 81
∴  a2 + b2 + c2 = 81 – 52 = 29
Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]
= 9[29 – 26]
= 9 x 3 = 27

Question 5.
If a + b + c = 9, and a2 + b2 + c2 = 35, find the value of a3 + b3 + c3 – 3abc.
Solution:
a + b + c = 9
Squaring, we get
(a + b + c)2 = (9)2
⇒  a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒  35 + 2(ab + bc + ca) = 81
2(ab + bc + ca) = 81 – 35 = 46
∴  ab + bc + ca = 462 = 23
Now, a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)]
= 9[35 – 23] = 9 x 12 = 108

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