RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.5 are available here. This exercise is about the sum and difference of cubes identity and conditional identity. RD Sharma Class 9 Maths PDF contains detailed answers prepared by the subject experts to help students study effectively, as well as prepare well to attempt all questions appearing in the exams. Students are advised to practise this RD Sharma chapter in order to ace their examinations.
RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.5
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 4 Algebraic Identities Exercise 4.5 Page Number 4.28
Exercise 4.5 Page No: 4.28
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) 2 – 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: If x + y + z = 8 and xy + yz+ zx = 20, find the value of x3 + y3 + z3 – 3xyz.
Solution:
We know, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Squaring, x + y + z = 8 both sides, 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 = 24
Now,
x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)]
= 8(24 – 20)
= 8 x 4
= 32
⇒ x3 + y3 + z3 – 3xyz = 32
Question 3: If a +b + c = 9 and ab + bc + ca = 26, find the value of a3 + b3 + c3 – 3abc.
Solution:
a + b + c = 9, ab + bc + ca = 26
Squaring, a + b + c = 9 both sides, we get
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2 (ab + bc + ca) = 81
a2 + b2 + c2 + 2 x 26 = 81
a2 + b2 + c2 + 52 = 81
a2 + b2 + c2 = 29
Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]
= 9[29 – 26]
= 9 x 3
= 27
⇒ a3 + b3 + c3 – 3abc = 27
RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.5
RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.5 are based on the following identities:
- Sum and difference of cubes identity
a3 + b3 + c3 – 3abc = (a+b+c)(a2 + b2 + c2 – ab – bc – ca)
- Conditional identity
If a + b+c = 0, then a3 + b3 + c3 = 3abc
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