Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.3 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions 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}

Answer:

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

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

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

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

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

= 9m^{2} + 30 m + 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} – 10 p + 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}

Answer:

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

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

(a + b)^{3} = a^{2} + 3a^{2}b + 3ab^{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} + 9m^{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) = a^{3} + 3a^{2}b + 3ab^{2} + b^{2}

(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}) – 4^{3}q^{3}

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

(iv) (52)^{3}

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

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

Answer:

(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 – 75x + 15 x^{2} – x^{3}

(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^{2} – 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 + 3ab (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}

(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}

(97xy)^{3} = 97^{3} x^{3} y^{3} = (100 – 3) x^{3}y^{3} … (1)

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 (p – 2)(p + 1)(p – 4)

Answer:

(p – 2)(p + 1)(p – 4) = (p + (-2)) (p + 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^{2} + (a + b + c) x^{2} + (ab + bc+ 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^{2} – 5p^{2} + 2p + 8

Question 5.

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

Answer:

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} = (a3^{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)

Answer:

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^{3} – 5x + 6 units^{3}

Objective Type Questions

Question 7.

If x^{2} – y^{2} = 16 and (x + y) = 8 then (x – y) is ________

(A) 8

(B) 3

(C) 2

(D) 1

Answer:

(C) 2

Hint:

x^{2} – y^{2} = 16

(x + y) (x – y) = 16

8 (x – y) = 16

(x – y) = \(\frac { 16 }{ 8 }\) = 2

Question 8.

\(\frac{(a+b)\left(a^{3}-b^{3}\right)}{\left(a^{2}-b^{2}\right)}\) = _________

(A) a^{2} – ab + b^{2}

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

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

(D) a2^{2} – 2ab + b^{2}

Answer:

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

Hint:

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

Question 9.

(p + q)(p^{2} – pq + q^{2}) is equal to __________

(A) p^{3} + q^{3}

(B) (p + q)^{3}

(C) p^{3} – q^{3}

(D) (p – q)^{3}

Answer:

(A) p^{3} + q^{3}

Hint:

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

Question 10.

(a – b) = 3 and ab = 5 then a^{3} – b^{3} = __________

(A) 15

(B) 18

(C) 62

(D) 72

Answer:

(D) 72

Hint:

(a – b) = 3

(a – b)^{2} = 3^{2}

a^{2} + b^{2} – 2ab = 9

a^{2} + b^{2} – 2(5) = 9

a^{2} + b^{2} = 9 + 10

a^{2} + b^{2} = 19

a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2}) = 3(19 + 5)

= 3(24) = 72

Question 11.

a^{3} + b^{3} = (a + b)^{3} _________

(A) 3a(a + b)

(B) 3ab(a – b)

(C) -3ab(a + b)

(D) 3ab(a + b)

Answer:

(D) 3ab(a + b)

Hint:

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

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

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