Haryana Board Class 8 Maths Algebra Chapter 7 cubes

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

Haryana Board Class 8 Maths Cubes and Cube Roots Solutions

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

Class 8 Maths Chapter 7 Cubes Haryana Board

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

Haryana Board 8th Class Maths Cube and Cube Root Questions and Answers

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

Chapter 7 Cubes Class 8 Solutions in Hindi Haryana Board

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

Haryana Board Class 8 Maths Exercise 7.1 Solutions

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

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

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

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

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

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

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

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

Important Questions for Class 8 Maths Chapter 7 Haryana Board

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).

Step-by-Step Solutions for Cubes Class 8 Haryana Board

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)