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**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) (9x^{2} + 4y^{2} + 4z^{2} – 6xy – 4yz – 6zx)

(ii) (4x -3y + 2z) (16x^{2} + 9y^{2}+ 4z^{2} + 12xy + 6yz – 8zx)

(iii) (2a – 3b – 2c) (4a^{2} + 9b^{2} + 4c^{2} + 6ab – 6bc + 4ca)

(iv) (3x -4y + 5z) (9x^{2} + 16y^{2} + 25z^{2} + 12xy- 15zx + 20yz)

Solution:

(i) (3x + 2y + 2z) (9x^{2} + 4y^{2} + 4z^{2} – 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

= 27x^{3} + 8y^{3} + 8Z^{3} – 36xyz

(ii) (4x – 3y + 2z) (16x^{2} + 9y^{2} + 4z^{2} + 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)

= 64x^{3} – 27y^{3} + 8z^{3} + 72xyz

(iii) (2a -3b- 2c) (4a^{2} + 9b^{2} + 4c^{2} + 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)

= 8a^{3} – 21b^{3} -8c^{3} – 36abc

(iv) (3x – 4y + 5z) (9x^{2} + 16y^{2} + 25z^{2} + 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)

= 27x^{3} – 64y^{3} + 125z^{3} + 180xyz

Question 2.

Evaluate:

Solution:

Question 3.

If x + y + z = 8 and xy + yz+ zx = 20, find the value of 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 – zx)

Now, x + y + z = 8

Squaring, we get

(x + y + z)^{2} = (8)^{2}x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) = 64

⇒ x^{2} + y^{2} + z^{2} + 2 x 20 = 64

⇒ x^{2} + y^{2} + z^{2} + 40 = 64

⇒ x^{2} + y^{2} + z^{2} = 64 – 40 = 24

Now,

x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) [x^{2} + y^{2} + z^{2} – (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 a^{3} + b^{3} + c^{3} – 3abc.

Solution:

a + b + c = 9, ab + be + ca = 26

Squaring, we get

(a + b + c)^{2} = (9)^{2}

a^{2} + b^{2} + c^{2} + 2 (ab + be + ca) = 81

⇒ a^{2} + b^{2} + c^{2} + 2 x 26 = 81

⇒ a^{2} + b^{2} + c^{2} + 52 = 81

∴ a^{2} + b^{2} + c^{2} = 81 – 52 = 29

Now, a^{3} + b^{3} + c^{3} – 3abc = (a + b + c) [(a^{2} + b^{2} + c^{2} – (ab + bc + ca)]

= 9[29 – 26]

= 9 x 3 = 27

Question 5.

If a + b + c = 9, and a^{2} + b^{2} + c^{2} = 35, find the value of a^{3} + b^{3} + c^{3} – 3abc.

Solution:

a + b + c = 9

Squaring, we get

(a + b + c)^{2} = (9)^{2}⇒ a^{2} + b^{2} + c^{2} + 2 (ab + be + ca) = 81

⇒ 35 + 2(ab + bc + ca) = 81

2(ab + bc + ca) = 81 – 35 = 46

∴ ab + bc + ca = 462 = 23

Now, a^{3} + b^{3} + c^{3} – 3abc

= (a + b + c) [a^{2} + b^{2} + c^{2} – (ab + bc + ca)]

= 9[35 – 23] = 9 x 12 = 108

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