Class 8 Maths

Cubes

Question 1. If P = 999, then find the volume of P (p²+3p+3).

Solution:

Given P = 999

= P(P²+3843)

= 999 (1999)²+3 (999)+3)

= 999(99800142997+3)

= 999 (1001001)

= 999,999,999

The volume of P (p²+3p+3) = 999,999,999

Question 2. Find the Cube of:

1. 1001

Solution:

Given: 1001

= (1001)³

– (1001)²(1001)

= 1002001 (1001)

= 1,003,003,001

(1001)³ = 1,003,003,001

2. 998

Solution:

Given: 998

= (998)³

= (998)²(992)

= 996004 (998)

= 994,001,1992

(998)³ = 994,001,1992

3. 5a-4b

Solution:

Given 5a-4b

= (5a-4b)³ [∵ (a-b)² = a³-3ab+3ab²+b³]

= (5a)³ – 3(5a)²4b + 3(5a)(4b)²-(4b)³

= 125a³ – 300a²b + 240ab² – 64b³

(5a-4b)³  = 125a³ – 300a²b + 240ab² – 64b³

4. 3a-b+4c

Solution:

= 3a-b+4c

= (3a-b+4c)³ [(∵(a+b+c)³ = a³ + b³ + c² + 3 (a+b) (b+c) (c+a)]

= (3a)³ + (-b)³ + (4c)³ + 3(3a-b)(-b+4c) (4c+3a)

= 27a³ – b³ + 64c³ + 3((3a-b) (-4bc – 3ab + 16c² + 12ac)

= 27a³-b³ + 64c³ +3[-12abc – 9a²b + 48ac² + 36²c + 4b²c + 3ab² – 16bc² = 12abc]

= 27a³-b³ + 64c³ – 36abc – 27a²b + 144ac² + 108a²C +12b²c + 9ab² – 48bc² – 36abc

= 27a³ – b³ + 64c³ – 27a²b + 9a² + 108a²c – 72abc + 12b²c +144ac² – 48bc²

(3a-b+4c)³= 27a³ – b³ + 64c³ – 27a²b + 9a² + 108a²c – 72abc + 12b²c +144ac² – 48bc²

Question 3. Simplify:

1. 3.2 x 3.2 x 3.2 – 3 x 3.2 x 3-2 x 1.2 + 3 x 3.2 × 1.2 x 1.2 −1·2 x 1.2 x 1.2

Solution:

Given:

= 3.2 x 32 x 3.2 – 3 x 3.2 x 3.2 x 1.2 + 3 x 3.2 x 1.2 x 1.2-1.2 x 1.2 x 1.2

= (3.2)³-3 (3-2)² x 1-2 + 3 x 3.2 x (1.2)² -(1.2)³

= 32.768 – 36.864 + 13.824 – 1.728

= 8

3.2 x 32 x 3.2 – 3 x 3.2 x 3.2 x 1.2 + 3 x 3.2 x 1.2 x 1.2-1.2 x 1.2 x 1.2 = 8

2. (a-b+c)³ + (a+b-c)³ + 6a [a²-(b-c)²]

Solution

Given:

= (a-b+c)³ + (a+b-c)³ + 6a [a²-(b-c)²]

= a³ – b³ + c³ + 3(a-b)(-b+ c)(c+a) + a³ + b² – c³ + 3(a+b)(b-c)(-c+a) + 6a [a²– b² + c² + 2bc]

= a³ – b³ + c³ + 3(a-b)(-bc-ba+c²+ca) + a³ + b³ – c³ + 3(a+b)(-bc+ba+c²-ca) + 6a [a²-b²+c²+2bc]

= a³ – b³ + c³ + 3 (-abc – a²b + ac² + a²c + b²c + b²a – bc² – abc) + a³ + b³ – c³ + 3(-abc + a²b + ac² – a²c -b²c + b²a + bc² – abc)+ 6a³ – 6ab² + 6ac² + 12abc

= a³– b³ + c³ – 3abc – 3a²b + 3ac²+ 3a²c + 3b²C + 3b²a – 3bc² – 3abc + a³ + b³ – c³ – 3abc + 3a²b + 3ac² – 3a²c – 3b²c + 3b²a + 3bc² – 3abc + 6ab³ – 6ab² + 6ac² + 12abc

= a³ – 3abc + 3ac² + 3b²a – 3abc – 3abc + 3ac² + 3b²a – 3abc + 6a³ – 6ab² + 6ac² + 12abc

= 8a³ – 12abc + 6ac² + 6ab² – 6ab² – 6ac² + 12abc

= 8a³

(a-b+c)³ + (a+b-c)³ + 6a [a²-(b-c)²] = 8a³

Question 4. If a-b = 8 then find the value of (a³-b³-24ab)

Solution:

Given: a-b=8

= a³-b³-24ab

= (a)³ – (b)³ -24ab

= (a-b)³ + 3 x a x b(a-b)- 24ab

= (8)³ +3ab (8)-24ab

= 512 + 24ab – 24ab

= 512

The value of (a³-b³-24ab) = 512

Question 5. Find the value of 8x³-36x²+54x-30 lf x=-5

Solution:

Given: X=-5

= 8x³-36x²+54x – 30

= 8(-5)³ – 36(-5)² + 54(-5)-30

= 8(-125)-36(25)-270-30

= -1000-900-270-30

= 1900+300

= 2200

The value of 8x³-36x²+54x-30 = 2200

Question 6. Find the product of the following:

1. (x-3)(x²+3x+9)

Solution:

Given: (x-3)(x²+3x+9)

= x³ + 3x² + 9x – 3x² – 9x – 27

= x³-27

(x-3)(x²+3x+9) = x³-27

2. (a²+b²)(a⁴-a²b²+b⁴)

Solution:

Given (a²+b²)(a⁴-a²b²+b⁴)

= (a²+b²)(a⁴-a²b²+b⁴)

= a²(a⁴-a²b²+b⁴) + b²(a⁴-a²b²+b⁴)

= a⁶ – a⁴b² + a²b⁴ + b²á⁴ – a²b⁴+ b⁶

= a⁶+b⁶

(a²+b²)(a⁴-a²b²+b⁴) = a⁶+b⁶

3. (4a²-9b²) (4a²+6ab+9b²) (4a²-6ab+9b²)

Solution:

Given: (4a²-9b²) (4a²+6ab+9b²) (4a²-6ab+9b²)

= (4a²(4a²+6ab+9b²)-9b²(4a²+6ab+9b²))(4a²-6ab+9b²)

= (16a⁴ + 24a³b + 36a²b² – 36a²b² – 54ab³ – 81(b⁴)(4a²-6ab+9b²)

= 64a⁶ + 96a⁵b – 216a³b³ – 324a²b⁴ – 96a⁵b – 144a⁴b² + 324a²b⁴ + 486ab⁵ + 144a⁴b² + 216a³b³ – 486ab⁵ – 729b⁶

= 64a⁶ – 729b⁶

(4a²-9b²) (4a²+6ab+9b²) (4a²-6ab+9b²) = 64a⁶ – 729b⁶

Question 7. Resolve into factors:

1. \(a^3-9 b^3-3 a b(a-b)\)

Solution:

Given: \(a^3-9 b^3-3 a b(a-b)\)

= \(a^3-9 b^3-3 a b(a-b)\)

= \(a^3-b^3-3 a b(a-b)-8 b^3\)

= \((a-b)^3-(2 b)^3\)

= \((a-b-2 b)\left\{(a-b)^2+(a-b) \cdot 2 b+(2 b)^2\right\}\)

= \((a-3 b)\left(a^2-2 a b+b^2+2 a b-2 b^2+4 b^2\right)\)

= \((a-3 b)\left(a^2+3 b^2\right)\)

\(a^3-9 b^3-3 a b(a-b)\) = \((a-3 b)\left(a^2+3 b^2\right)\)

2. a¹² – b¹²

Solution:

Given a¹² – b¹²

= (a⁶)² – (b⁶)²

= \(\left(a^6+b^6\right)\left(a^6-b^6\right)\)

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

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

= \(\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)(a+b)\left(a^2-a b+b^2\right)(a-b)\left(a^2+a b+b^2\right)\)

= \((a+b)(a-b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)\)

a¹² – b¹² = \((a+b)(a-b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)\)

Question 8. Choose the Correct answer:

1. The Cube of 99 is

1. 972099
2. 970299
3. 979029
4. 972909

Solution:

=(99)³

= (99)² 99

= 9801 x 99

= 970299

(99)³ = 970299

The Correct answer is (2).

2. If a+b+c=0, then a³+b³+c³ = ?

1. 3abc
2. abc
3. c
4. none of these

Solution:

= a³ + b³ + c³

= (a+b)³ -3ab (a+b) + c³

= (-c)³-3ab (-c) +c³ [a+b+c=0)

= -c³+3abc+c³

= 3abc

a³ + b³ + c³ = 3abc

So the Correct answer is (1).

3. If a-b=1 and a³-b³=61, then the Value of ab is

1. 10
2. 20
3. 30
4. none of these

Solution:

= a³-b³=61

= (a-b)³ + 3ab (a-b) = 61

= (1)³ + 3ab(1) = 61

= 3ab = 61-1

= 3ab = 60

= ab = \(\frac{60}{3}\)

= ab = 20

So, the Correct answer is (2).

Question 9. write ‘True’ or ‘False”.

1. 216 is not a perfect Cube.

Solution:

216 = 36×6

216= 6x6x6

216=(6)³

The statement is False.

2. 1729 is a Hardy Ramanujam number.

Solution: The statement is true.

3. p³q³+1 = (pq-1)(p²q² + pq + 1)

Solution:

p³q³ + 1 = (pq)³ + (1)³

p³q³ + 1= (pq+1){(pq)² – pq.1 + (1)2}

p³q³ + 1= (pq+1) (p²q² – pq+1)

Statement is False.

Question 10. Fill in the blanks.

1. The Cube root of Single digit number may also be a _________ digit number.

Solution: Single.

2. a³+b³ = ___________ x (a²-ab+b²)

Solution: a³ + b³ = (a+b) (a²-ab+b²)

3. (a+b)³ = a² + b³ + __________.

Solution: (a+b)³ = a³ + b³ + 3ab(a+b)

Factorisation Of Algebraic Expressions

Question 1. Resolve into factors:

1. x² – 22x + 120

Solution:

Given: x² – 22x + 120

= x²-10x-12x-120

= x(x-10)-12(x-10)

= (x-10) (x-12)

x² – 22x + 120 = (x-10) (x-12)

2. 40+3x-x²

Solution:

Given: 40+3x-x²

= 40+8x-5x-x²

= 8(5+x)-x(5+x)

= (8-x)(5+x)

40+3x-x² = (8-x)(5+x)

3. (a²-a)²-8(a²-a)+12

Solution:

Given: (a²-a)²-8(a²-a)+12

= (a²-a)² – 6(a²-a)-2(a²-a)+12

= (a²−a)((a²-a)−6)-2((a²-a)-6)

= (a²-a-2) (a²-a-6)

= (a²+a-2a-2)(a²3a+2a-6)

= (a(a+1)-2(a+1))(a(a-3)+2(9-3))

= (a+1) (a-2) (a+2) (a-3)

(a²-a)²-8(a²-a)+12 = (a+1) (a-2) (a+2) (a-3)

4. x²-√3x-6

Solution:

Given: x²-√3x-6

= x²-2√3x+√3x-6

= x(x-2√3)+ √3(x-2√3)

= (x+√3)(x-2√3)

x²-√3x-6 = (x+√3)(x-2√3)

5. (x+1)(x+3)(x-4)(x-6)+13

Solution:

Given (x+1)(x+3)(x-4)(x-6)+ 13

= (x²+3x+x+3)(x²-6x-4x+24)+13

= (x²+4x+3)(x²-10x+24)+13

= x⁴-10x³+24x²+4x³40x²+96x+3x² = 30x+72+13

= x⁴-6x³-13x²-66+85

= x⁴-3x³-3x³-17x²+9x²– 5x²+51x+15x+85

= x⁴-3x³-17x²-3x³+9x²+51x-5x²+15x+85

= x²(x²-3x-17)-3x(x²-3x-17)-5(x²-3x-17)

= (x²-3x-5)(x²-3x-17)

(x+1)(x+3)(x-4)(x-6)+ 13 = (x²-3x-5)(x²-3x-17)

6. 21x² + 40xy – 21y²

Solution:

Given: 21x²+40xy-21y²

= 21x² – 9xy + 49xy – 21y²

= 3x(7x-3y) + 7y(7x-3y)

= (3x+7y)(7x-3y)

21x²+40xy-21y² = (3x+7y)(7x-3y)

7. 4(2a-3)² -3(2a-3) (a-1)-7(a-1)²

Solution:

Given: 4(2a-3)²-3(2a-3) (a-1)-7(a-1)²

= 4(4a²-(2a+9)-3(2a²-2a-3a+3)-7(a²+1-2a)

= 16a² – 48a + 36 – 6a² + 6a + 9a – 9 – 7a² – 7 + 14

= 3a² – 19a + 20

= 3a² – 15a – 4a +20

= 3(a-5)-4(a-5)

4(2a-3)²-3(2a-3) (a-1)-7(a-1)² = (3a-4) (a-5)

8. (a+7) (a-10) + 16

Solution:

Given: (a+7)(a-10)+16

= a²-10a+7a-70+16

= a²-3a-54

= a-9a+6a-54

= a(a-9)+6(a-9)

=(a+6)(a-9)

(a+7)(a-10)+16 =(a+6)(a-9)

9. (a²+4a)²+21(a²+4a)+98

Solution:

Given: (a²+4a)²+21(a²+4a)+98

= a⁴ + 8a³ + 16a² + 2(a²+4a)+ 98

= a⁴ + 8a³ + 37a² +84a +98

= a⁴+4a³+14a²+4a³+16a²+56a+7a²+28+98

= a² (a²+4a+14)+4a(a²+4a+14) +7(a²+4a+14)

= (a²+4a+7) (a² + 4a + 14)

(a²+4a)²+21(a²+4a)+98 = (a²+4a+7) (a² + 4a + 14)

10. (2a²+ 5a) (2a² + 5a-19) +84

Solution:

Given: (2a²+ 5a) (2a² + 5a-19) +84

= 4a⁴ + 10a³ – 38a² + 10a³ + 25a² – 95a + 84

= 4a⁴ + 20a³ – 13a² – 95a + 84

= 4a⁴ + 8a³ – 21a² + 12a³ + 24a² – 63a -16a² – 32a + 84

= a²(4a²+8a-21)+3a(4a²+8a-21)-4(4a²+8a-21)

= (a²+3a-4) (4a² +8a-21)

= (a²+4a-a-4) (4a² + 14a-6a-21)

= ((a+4)a-1(a+4)) (2a(2a+7)-3(2a+7))

= (a-1) (a+4) (2a-3)(2a+7)

(2a²+ 5a) (2a² + 5a-19) +84 = (a-1) (a+4) (2a-3)(2a+7)

11. 8x⁴+2x²-45

Solution:

Given: 8x⁴+2x²-45

= 8x² + 20x² – 18x² – 45

= 4x² (2x²+5)-9(2x²+5)

= (2x²+5)(4x²-9)

= (2x+5) (4x²-6x+6x-9)

= (2x²+5) (2x(2x-3)+3(2x-3))

= (2x²+5)(2x+3)(2x-3)

8x⁴+2x²-45 = (2x²+5)(2x+3)(2x-3)

12. xv-x-(a-3)(a-2)

Solution:

Given: x²-x-(a-3)(a-2)

= x²-x-(a²-2a-3a+6)

= x²-x-a²+5a-6

= x²-3x+2x-a²+ax-ax+3a+2a-6

= x²+ax-3x-ax-a²+3a+2x+2a-6

= x(x+a-3)-a(x+a-3)+2(x+a-3)

=(x-a+2)(x+a-3)

x²-x-(a-3)(a-2) =(x-a+2)(x+a-3)

13. 99x²-20xy+99y²

Solution:

Given: 99x²-20xy+99y²

= 99x²-81xy-121xy+99y²

= 9x (11x-9y)-11y (11x-94)

= (9x-114) (11x-94)

99x²-20xy+99y²  = (9x-114) (11x-94)

Question 2. Resolve the following expressions into factors by expressing them as the diffence of two Squares:

1. x²-5x-6

Solution:

Given: x²-5x-6

= x²-6x+x-6

= x(x-6)+1(x-6)

= (x+1)(x-6)

x²-5x-6 = (x+1)(x-6)

2. 3+x-10x²

Solution:

Given: 3+x-10x²

= 3+6x-5x-10x²

= 3(1+2x)-5x(1+2x)

= (3-5x) (1+2x)

3+x-10x² = (3-5x) (1+2x)

3. 8x-3-4x²

Solution:

Given: 8x-3-4x²

= 6x+2x-3-4x²

= 6x-3+2x-4x²

= 3(2x-1)-2x(2x-1)

= (3-2x) (2x-1)

8x-3-4x²= (3-2x) (2x-1)

4. 6(a+b)²+5(a²-b²)-6 (a-b)²

Solution:

Given: 6(a+b)²+5(a²-b²)-6(a-b)²

= 6(a²+b²+2ab)+5a²-5b²-6(a²+b²-2ab)

= 6a²+6b²+ 12ab+ 5a²-5b²-6a²-6b² +12ab

= 5a²+24ab-5b²

= 5a² +25ab-ab-5b²

= 5a(a+5b)-b(a+5b)

= (5a-b)(a+5b)

6(a+b)²+5(a²-b²)-6(a-b)² = (5a-b)(a+5b)

5. 6x²-13x+6

Solution:

Given: 6x²-13x+6

= 6x²-9x-4x+6

= 3x(2x-3)-2(2x-3)

=(3x-2)(2x-3)

6x²-13x+6 =(3x-2)(2x-3)

Question 3. Choose the Correct answer:

1. x²-3x-28 = ?

1. (x+4)(x+7)
2. (x+4)(x-7)
3. (x-4)(x+7)
4. (x-4)(x-7)

Solution:

x²-3x-28

= x²-7x+4x-28

= x(x-7)+4(x-7)

= (x+4) (x-7)

x²-3x-28 = (x+4) (x-7)

The Correct answer is (2)

2) If (5x²-4x-9) = (x+1)(5x+P), then the value of P is

1. 9
2. 5
3. -9
4. none of these

Solution:

= 5x²-4x-9

= 5x²-9x+5x-9 = x(5x-9)+1(5x-9)

= (5x-9)(x+1)

= (x+1)(5x+P)= (x+1) (5x-9)

= 5x + P = 5x-9

⇒ P = -9

The value of P = -9

The Correct answer is (3).

3. 2a²+b²-c²+3ab+ac = ?

1. (a+b+c)(2a+b+c)
2. (a+b+c)(2a+b-c)
3. (a+b+c) (2a-b-c)
4. none of these

Solution:

2a²+b²-c²+3ab+ac

= 2a²+ab-ac+2ab+b²-bc+2ac + bc-c²

= a(2a+b-c)+b(2a+b-c) + C(2a+b-c)

= (a+b+c) (2a+b-c)

2a²+b²-c²+3ab+ac = (a+b+c) (2a+b-c)

The Correct answer is (2).

Question 4. write ‘True’ or ‘False’:

1. The Factors of (x²-xy-30y²) is (x+5y)(x-64).

Solution:

x²-xy-30y² = x²-6xy+5xy-30y²

= x(x-6y) +5y(x-6y) = (x-6y) (x+5y)

The statement is true.

2. a³-b³-a(a²-b²)+b(a-b)²=ab(a-b)

Solution:

a³-b³-a(a²-b²)+b(a-b)²=ab(a-b)

= a³-b³-a³+ab²+b(a²+b²-2ab)

= -b³+ab²+ba²+b³-2ab²

= a²b-ab² = ab(a-b)

The statement is true.

3.(x-1)(x+9)+21=(x+6)(x-2)

Solution:

(x-1)(x+9)+21

= x²+9x-x-9+21

= x²+8x+12

= x²+6x+2x+12

= x(x+6)+2(x+6)

= (x+2)(x+6)

The statement is false.

Question 5. Fill in the blanks:

1. (x+a)(x+b) = x²+(a+b)x + _______.

Solution:

(x+9)(x+6) = x(x+b)+a(x+b)

= x²+bx+ax+ab = x²+(a+b)x+ab

2. (a+b+c)³ = a³ + b² + c² + __________.

Solution:

3(a+b)(b+c)(c+a).

• Chapter 1 Rational Numbers
• Chapter 2 Linear Equation in One Variable
• Chapter 3 Understanding Quadrilaterals
• Chapter 4 Practical Geometry
• Chapter 5 Data Handling
• Chapter 6 Square and Square Roots
• Chapter 7 Cube and Cube Roots
• Chapter 8 Comparing Quantities
• Chapter 9 Algebraic Expressions and Identities
• Chapter 10 Visualising Solid Shapes
• Chapter 11 Mensuration
• Chapter 12 Exponents and Powers
• Chapter 13 Direct and Indirect Proportions
• Chapter 14 Factorisation
• Chapter 15 Introduction to Graphs
• Chapter 16 Playing with Numbers