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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

Question 1.

Expand

(i) (3m + 5)^{2}

(ii) (5p – 1)^{2}

(iii) (2n – 1)(2n + 3)

(iv) 4p^{2} – 25q^{2}

Solution:

(i) (3m + 5)^{2}

Comparing (3m + 5)^{2} with (a + b)^{2} we have a = 3m and b = 5

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(3m + 5)^{2} = (3m)^{2} + 2 (3m) (5) + 5^{2}

= 3^{2}m^{2} + 30m + 25 = 9m^{2} + 30m +25

(ii) (5p – 1)^{2}

Comparing (5p – 1)^{2} with (a – b)^{2} we have a = 5p and b = 1

(a – b)^{2} = a^{2} – 2ab + b^{2}

(5p – 1)^{2} = (5p)^{2} – 2 (5p) (1) + 1^{2}

= 5^{2}p^{2} – 10p + 1 = 25p^{2} – 10p + 1

(iii) (2n – 1)(2n + 3)

Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3

(x + a) (x + b) = x^{2} + (a + b)x + ab

(2n +(- 1)) (2n + 3) = (2n)^{2} + (-1 + 3)2n + (-1) (3)

= 2^{2}n^{2} + 2 (2n) – 3 = 4n^{2} + 4n – 3

(iv) 4p^{2} – 25q^{2} = (2p)^{2} – (5q)^{2}

Comparing (2p)^{2} – (5q)^{2} with a^{2} – b^{2} we have a = 2p and b = 5q

(a^{2} – b^{2}) = (a + b)(a – b) = (2p + 5q) (2p – 5q)

Question 2.

Expand

(i) (3 + m)^{3}

(ii) (2a + 5)^{3}

(iii) (3p + 4q)^{3}

(iv) (52)^{3}

(v) (104)^{3}

Solution:

(i) (3 + m)^{3}

Comparing (3 + m)^{3} with (a + b)^{3} we have a = 3; b = m

(a + b)^{3} = a^{2} + 3a^{2}b + 3 ab^{2} + b^{3}

(3 + m)^{3} = 3^{3} + 3(3)^{2} (m) + 3 (3) m^{2} + m^{3}

= 27 + 27m + 9m^{2} + m^{3} = m^{3} + 9 m^{2} + 27m + 27

(ii) (2a + 5)^{3}

Comparing (2a + 5)^{3} with (a + b)^{3} we have a = 2a, b = 5

(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} = (2a)^{3} + 3(2a)^{2} 5 + 3 (2a) 5^{2} + 5^{3}

= 2^{3}a^{3} + 3(2^{2}a^{2}) 5 + 6a (25) + 125

= 8a^{3}+ 60a^{2} + 150a + 125

(iii) (3p + 4q)^{3}

Comparing (3p + 4q)^{3} with (a + b)^{3} we have a = 3p and b = 4q

(a + b) ^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

(3p + 4q)^{3} = (3p)^{3} + 3(3p)^{2} (4q) + 3(3p)(4q)^{2} + (4q)^{3}

= 3^{3}p^{3} +3 (9p^{2}) (4q) + 9p (16q^{2}) + 43q^{3}

= 27p^{3} + 108p^{2}q + 144pq^{2} + 64q^{3}

(iv) (52)^{3} = (50 + 2)^{3}

Comparing (50 + 2)^{3} with (a + b)^{3} we have a = 50 and b = 2

(a + b)^{3} = a^{3} + 3 a^{2}b + 3 ab^{2} + b^{3}

(50 + 2)^{3} = 50^{3} + 3 (50)^{2}2 + 3 (50)(2)^{2} + 2^{3}

52^{3} = 125000 + 6(2,500) + 150(4) + 8

= 1,25,000 + 15,000 + 600 + 8

52^{3} = 1,40,608

(v) (104)^{3} = (100 + 4)^{3}

Comparing (100 + 4)^{3} with (a + b)^{3} we have a = 100 and b = 4

(a + b)^{3} = a^{3} + 3 a^{2}b + 3 ab^{2} + b^{3}

(100 + 4)^{3} = (100)^{3} + 3 (100)^{2} (4) + 3 (100) (4)^{2} + (4)^{3}

= 10,00,000 + 3(10000) 4 + 300 (16) + 64

= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

Question 3.

Expand

(i) (5 – x)^{3}

(ii) (2x – 4y)^{3}

(iii) (ab – c)^{3}

(iv) (48)^{3}

(v) (97xy)^{3}

Solution:

(i) (5 – x)^{3}

Comparing (5 – x)^{3} with (a – b)^{3} we have a = 5 and b = x

(a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3}

(5 – x)^{3} = 5^{3} – 3 (5)^{2} (x) + 3(5)(x^{2}) – x^{3}

= 125 – 3(25)(x) + 15x^{2} – x^{3} = 125

(ii) (2x – 4y)^{3}

Comparing (2x – 4y)^{3} with (a – b)^{3} we have a = 2x and b = 4y

(a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{3} – b^{3}

(2x – 4y)^{3} = (2x)^{3} – 3(2x)^{2} (4y) + 3(2x) (4y)^{2} – (4y)^{3}

= 2^{3}x^{3} – 3(2^{2}x^{2}) (4y) + 3(2x) (4^{2}y^{2}) – (4^{3}y^{3})

= 8x^{3} – 48x^{2}y + 96xy^{2} – 64y^{3}

(iii) (ab – c)^{3}

Comparing (ab – c)^{3} with (a – b)^{3} we have a = ab and b = c

(a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3}

(ab – c)^{3} = (ab)^{3} – 3 (ab)^{2} c + 3 ab (c)^{2} – c^{3}

= a^{3}b^{3} – 3(a^{2}b^{2}) c + 3abc^{2} – c^{3}

= a^{3}b^{3} – 3a^{2}b^{2} c + 3abc^{2} – c^{3}

(iv) (48)^{3} = (50 – 2)^{3}

Comparing (50 – 2)^{3} with (a – b)^{3} we have a = 50 and b = 2

(a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3}

(50 – 2)^{3} = (50)^{3} – 3(50)^{2}(2) + 3 (50)(2)^{2} – 2^{3}

= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592

= 1,10,592

(v) (97xy)^{3}

= 97^{3} x^{3} y^{3} = (100 – 3)^{3} x^{3}y^{3}

Comparing (100 – 3)^{3} with (a – b)^{3} we have a = 100, b = 3

(a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3}

(100 – 3)^{3} = (100)^{3} – 3(100)^{2} (3) + 3 (100)(3)^{2} – 3^{3}

97^{3} = 10,00,000 – 90000 + 2700 – 27

97^{3} = 910000 + 2673

97^{3} = 912673

97x^{3}y^{3} = 912673x^{3}y^{3}

Question 4.

Simplify (i) (5y + 1)(5y + 2)(5y + 3)

(ii) (p – 2)(p + 1)(p – 4)

Solution:

(i) (5y + 1) (5y + 2) (5y + 3)

Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.

(x + a) (x + b) (x + c) = x^{3} + (a + b + c) x^{2} (ab + bc + ca) x + abc

= (5y)^{3} + (1 + 2 + 3) (5y)^{2} + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)

= 5^{3}y^{3} + 6(5^{2}y^{2}) + (2 + 6 + 3)5y + 6

= 125^{3} + 150y^{2} + 55y + 6

(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))

Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.

(x + a) (x + b) (x + c) = x^{3} + (a + b + c) x^{2} + (ab + be + ca) x + abc

= p^{3} + (-2 + 1 + (-4))p^{2} + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)

= p^{3} + (-5 )p^{2} + (-2 + (-4) + 8)p + 8

= p^{3} – 5p^{2} + 2p + 8

Question 5.

Find the volume of the cube whose side is (x + 1) cm.

Solution:

Given side of the cube = (x + 1) cm

Volume of the cube = (side)^{3} cubic units = (x + 1)^{3} cm^{3}

We have (a + b)^{3} = (a^{3} + 3a^{2}b + 3ab^{2} + b^{3}) cm^{3}

(x + 1)^{3} = (x^{3} + 3x^{2} (1) + 3x (1)^{2} + 1^{3}) cm^{3}

Volume = (x^{3} + 3x^{2} + 3x + 1) cm^{3}

Question 6.

Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3).

Solution:

Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)

∴ Volume of the cuboid = (l × b × h) units^{3}

= (x + 2) (x – 1) (x – 3) units^{3}

We have (x + a) (x + b) (x + c) = x^{3} + (a + b + c) x^{2} + (ab + bc+ ca)x + abc

∴ (x + 2)(x – 1) (x – 3) = x^{3} + (2 – 1 – 3)x^{2} + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)

x^{3} – 2x^{2} + (-2 + 3 – 6)x + 6

Volume = x^{3} – 2x^{2} – 5x + 6 units^{3}