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ICSE • कक्षा 8 • Mathematics • अध्याय 5
अनुमानित समय: 45 Mins
प्रगति: अध्ययनरत

घन और घनमूल (Cubes and Cube Roots)

In ICSE Class 8 Mathematics, "Cubes and Cube Roots" provides an authoritative, mathematically rigorous master study guide investigating perfect cubes, properties of cubic numbers, extraction of cube roots by prime factorisation, estimation method for perfect cubes, and cube roots of negative numbers, fractions, and decimals. This comprehensive chapter explores Definition of a Perfect Cube ($n = m^3 = m \times m \times m$ for $m \in \mathbb{Z}$; Table of cubes of the first 20 natural numbers), Properties of Cubes (Cubes of even numbers are always even; Cubes of odd numbers are always odd; Cubes of negative integers are always negative: $(-a)^3 = -a^3$; Cube of numbers ending in digits: $1 \to 1, 2 \to 8, 3 \to 7, 4 \to 4, 5 \to 5, 6 \to 6, 7 \to 3, 8 \to 2, 9 \to 9, 0 \to 000$; Sum of cubes identity: $1^3 + 2^3 + 3^3 + \dots + n^3 = (1 + 2 + 3 + \dots + n)^2$), Methods of Extracting Cube Roots: 1. Prime Factorisation Method (grouping identical factors into triplets of three: $\sqrt[3]{a^3 b^3} = ab$), 2. Estimation Method for Large Perfect Cubes (grouping into triads from right to left; unit's digit determines unit's place, remaining group determines ten's digit), Cube Roots of Products, Rational Fractions, and Decimals ($\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$ and $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$), Smallest Multiplier or Divisor to Make a Number a Perfect Cube, and Practical Volume Applications ($V = s^3 \implies s = \sqrt[3]{V}$) aligned with the 2026–27 CISCE ICSE curriculum.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें
प्रस्तावना अध्याय अवधारणाएँ सूत्र एवं सर्वसमिकाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

How Did an Uneducated 32-Year-Old Indian Mathematical Genius Memorize the Secret Cubic Soul of the Number 1729 on His Deathbed?

In 1918, in a cold nursing home in Putney, England, the great English mathematician G.H. Hardy walked into the room to visit his dying protégé, the legendary Indian genius Srinivasa Ramanujan. Hardy remarked offhandedly: "I rode here in taxi-cab number 1729. It seemed to me a rather dull, uninteresting number." Ramanujan instantly sat bolt upright in bed, eyes blazing, and replied: "No, Hardy! It is a very interesting number! 1729 is the smallest number expressible as the sum of two cubes in two different ways!" Ramanujan instantly visualized that: $1729 = 1^3 + 12^3 = 1 + 1728$, and $1729 = 9^3 + 10^3 = 729 + 1000$! Today, 1729 is celebrated across the world as the Hardy-Ramanujan Number! While a square ($s^2$) represents flat two-dimensional surface area, a Cube ($s^3$) captures the three-dimensional volume of our physical space! Why is the cube root of a negative number always negative? How can you look at a six-digit number like $175,616$ and calculate its cube root in two seconds in your head using Estimation? Let's master cubes and cube roots.

यह अध्याय क्यों महत्वपूर्ण है

Cubic calculations govern 3D spatial volumes, fluid tank capacities in chemical engineering, planetary gravity densities, stress-strain tensor mechanics, and crystal lattice nanotechnology. Mastering prime triplet grouping and estimation is a fundamental requirement in ICSE Class 8 algebra.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Squares and square roots from Chapter 4.
  • Prime factorisation using division.
  • Operations on negative integers.

इस अध्याय के लक्ष्य

  • Identify perfect cubes and understand unit-digit terminal relationships.
  • Find the cube root of integers using the Prime Factorisation Triplet Method.
  • Extract the cube root of negative integers, fractions, and decimals ($\\sqrt[3]{-a} = -\\sqrt[3]{a}$).
  • Determine the cube root of large perfect cubes rapidly using the Estimation Triad Method.
  • Find the smallest number by which a given number must be multiplied or divided to become a perfect cube.
  • Solve real-world volume word problems involving cubic dimensions.

अध्याय रूपरेखा एवं प्रगति

1 1. Concept of Cubes & Unit Digit Pa...
2 2. Properties of Cubes & Negative C...
3 3. Prime Factorisation & The Estima...
4 4. Cube Roots of Fractions, Decimal...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Concept of Cubes & Unit Digit Patterns

Understand
A. What is a Cube?

The Cube of a number is the product obtained by multiplying the number by itself three times:

$$\mathbf{n = m^3 = m \times m \times m}$$
  • The inverse operation of finding the cube is called extracting the Cube Root (symbol: $\mathbf{\sqrt[3]{\phantom{x}}}$). $$m^3 = n \iff m = \sqrt[3]{n}$$
B. Terminal Unit Digit Invariance:
Number Ends In Cube Ends In Rule / Complement
0, 1, 4, 5, 6, 90, 1, 4, 5, 6, 9Ends in the exact same digit!
28Complements adding to 10 ($2 + 8 = 10$)
82Complements adding to 10 ($8 + 2 = 10$)
37Complements adding to 10 ($3 + 7 = 10$)
73Complements adding to 10 ($7 + 3 = 10$)

2. Properties of Cubes & Negative Cube Roots

Properties
  1. Parity:
    • The cube of an even number is always even: $4^3 = 64, 6^3 = 216$.
    • The cube of an odd number is always odd: $3^3 = 27, 5^3 = 125$.
  2. Negative Numbers (Odd Power):

    Unlike square roots (which cannot be negative for real numbers), the cube of a negative number is always negative, and the cube root of a negative number is always negative:

    $$\mathbf{(-a)^3 = -a^3 \quad \Longleftrightarrow \quad \sqrt[3]{-x} = -\sqrt[3]{x}}$$ Example: $\sqrt[3]{-216} = -\sqrt[3]{216} = \mathbf{-6}$.
  3. Terminal Zeros: A cube number always ends in a multiple of three zeros ($3, 6, 9$ zeros). If a number ends in $1$ or $2$ zeros, it can NEVER be a perfect cube.

3. Prime Factorisation & The Estimation Method

Extraction Methods
Method 1: Prime Factorisation (Grouping in Triplets):

Resolve the number into prime factors, group identical factors into triplets of three, and take one factor from each triplet:

$$\sqrt[3]{216} = \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)} = 2 \times 3 = \mathbf{6}$$
Method 2: Rapid Estimation for Perfect Cubes:

To find the cube root of a large perfect cube (e.g., $175,616$):

  1. Form groups of three digits (triads) starting from the right: Group 1 (units group): $616$; Group 2 (ten's group): $175$.
  2. Unit's Digit: Group 1 ends in $6$. Since only $6^3$ ends in $6$, the unit's digit of the cube root must be $6$.
  3. Ten's Digit: Group 2 is $175$. Find between which two perfect cubes it lies: $$5^3 = 125 < 175 < 6^3 = 216$$ The smaller cube is $5^3$, so the ten's digit is $5$.
  4. Combining digits gives: $$\mathbf{\sqrt[3]{175616} = 56}$$ !

4. Cube Roots of Fractions, Decimals & Smallest Multipliers

Fractions & Multipliers
A. Product & Quotient Rules:
$$\mathbf{\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \quad \land \quad \sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}}$$

Example: $\sqrt[3]{0.027} = \sqrt[3]{\frac{27}{1000}} = \frac{\sqrt[3]{27}}{\sqrt[3]{1000}} = \frac{3}{10} = \mathbf{0.3}$.

B. Smallest Multiplier or Divisor Problems:

Resolve the number into prime factors:

  • To find the least multiplier to make it a perfect cube: supply the missing factors needed to complete incomplete triplets.
  • To find the least divisor: divide by the leftover factors that do not form complete triplets.

महत्वपूर्ण सूत्र, सर्वसमिकाएँ एवं प्रमेय

Negative Cube Root Identity
$$\sqrt[3]{-a} = -\sqrt[3]{a}$$
Cube root of negative real number is strictly negative.
Sum of Cubes Formula
$$1^3 + 2^3 + 3^3 + \dots + n^3 = \left[ \frac{n(n+1)}{2} \right]^2$$
The sum of the first n cubes equals the square of their sum.

Cubic Geometry: Unit Digit Complements & Estimation Triads

Cubes & Cube Roots: Unit Digit Rules & Estimation Triads UNIT DIGIT TRANSFORMATION RULES • Digits that REMAIN IDENTICAL in Cube: 0 → 0 • 1 → 1 • 4 → 4 • 5 → 5 • 6 → 6 • 9 → 9 • The "Sum to 10" Complement Pairs: 2 ↔ 8 (23=8, 83=512) 3 ↔ 7 (33=27, 73=343) • Negative Cubes: √[3](-x) = -√[3](x) • √[3](-216) = -6 • Cubes must end in multiples of 3 zeros (000) ESTIMATION METHOD: √[3](175616) Group 2: 175 (Ten's Digit) Group 1: 616 (Unit's Digit) Step 1: Unit's Place from 616: Number ends in 6 ⇒ Unit's digit = 6 Step 2: Ten's Place from 175: 53 = 125 < 175 < 63 = 216 ⇒ Ten's digit = 5 √[3](175616) = 56 CUBE ROOT OF NEGATIVE IS NEGATIVE • PAIRS: 2↔8 AND 3↔7 • GROUP IN TRIADS OF THREE

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
A perfect cube is formed by multiplying a number by itself three times (n = m^3).
मुख्य बिंदु 2
Cubes of even numbers are even; cubes of odd numbers are odd.
मुख्य बिंदु 3
The cube and cube root of a negative integer are always negative: (-a)^3 = -a^3.
मुख्य बिंदु 4
Digits 0, 1, 4, 5, 6, 9 retain their identity in cubes; 2 pairs with 8, and 3 pairs with 7.
मुख्य बिंदु 5
A perfect cube must end in a multiple of three zeros (000, 000000).
मुख्य बिंदु 6
Cube root by prime factorisation requires grouping factors into triplets of three identical primes.
मुख्य बिंदु 7
The estimation method rapidly finds cube roots of large perfect cubes by grouping in triads.
मुख्य बिंदु 8
Cube root of a product: cuberoot(a*b) = cuberoot(a) * cuberoot(b).
मुख्य बिंदु 9
Cube root of a fraction: cuberoot(a/b) = cuberoot(a) / cuberoot(b).
मुख्य बिंदु 10
Volume of a cube is V = s^3; side length is the cube root of the volume (s = cuberoot(V)).

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
Find the cube root of $-17576$ using the Prime Factorisation Method.
उत्तर एवं व्याख्या देखें
उत्तर: Step 1: Using the negative cube root property: $\sqrt[3]{-17576} = -\sqrt[3]{17576}$.
Step 2: Prime factorise $17576$:
• $17576 \div 2 = 8788$
• $8788 \div 2 = 4394$
• $4394 \div 2 = 2197$
• $2197 \div 13 = 169$
• $169 \div 13 = 13$
• $13 \div 13 = 1$
$$17576 = (2 \times 2 \times 2) \times (13 \times 13 \times 13)$$
Step 3: Group into triplets and take one factor from each triplet:
$$\sqrt[3]{17576} = 2 \times 13 = 26$$
Step 4: Incorporate the negative sign:
$$\mathbf{\sqrt[3]{-17576} = -26}$$.
Factorise 17576 into $2^3 \times 13^3$. Cube root is $2 \times 13 = 26$. Include negative sign: $-26$.
2
Find the smallest number by which $675$ must be multiplied so that the product becomes a perfect cube.
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Prime factorise $675$:
• $675 \div 5 = 135$
• $135 \div 5 = 27$
• $27 \div 3 = 9$
• $9 \div 3 = 3$
• $3 \div 3 = 1$

$$675 = (3 \times 3 \times 3) \times (5 \times 5)$$


Step 2: Examine the prime factor triplets:
• The prime factor $3$ forms a complete triplet: $(3 \times 3 \times 3)$.
• The prime factor $5$ appears only twice: $(5 \times 5)$—it lacks one more 5 to complete a triplet.
• Therefore, the smallest number by which $675$ must be multiplied is $5$.
(Check: $675 \times 5 = 3375 = 15^3$).


Prime factorise: $3^3 \times 5^2$. It needs one more 5 to complete the triplet of 5s.
3
Find the smallest number by which $8640$ must be divided so that the quotient is a perfect cube.
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Prime factorise $8640$:
• $8640 = 2 \times 4320 = 2^2 \times 2160 = 2^3 \times 1080 = 2^4 \times 540 = 2^5 \times 270 = 2^6 \times 135$
• $135 = 3 \times 45 = 3^2 \times 15 = 3^3 \times 5$

$$8640 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times 5 = 2^6 \times 3^3 \times 5$$


Step 2: Group into triplets:
• $2^3, 2^3,$ and $3^3$ all form complete triplets.
• The factor $5$ is leftover and cannot form a triplet.
• Therefore, the smallest number by which $8640$ must be divided is $5$.
(Check: $8640 \div 5 = 1728 = 12^3$).


$8640 = 2^6 \times 3^3 \times 5$. The leftover factor is 5, so divide by 5 to get $1728 = 12^3$.
4
Using the Estimation Method, find the cube root of the large perfect cube $493,039$.
उत्तर एवं व्याख्या देखें
उत्तर:

Step 1: Split the number into two triads (groups of 3 digits) from right to left:
Group 1 (right): $\mathbf{039}$; Group 2 (left): $\mathbf{493}$.

Step 2: Determine the Unit's Digit from Group 1 ($039$):
The number ends in $9$. Only $9^3 = 729$ ends in $9$.
Therefore, the unit's digit of the cube root is $9$.

Step 3: Determine the Ten's Digit from Group 2 ($493$):
Find the two consecutive perfect cubes between which $493$ lies:

$$7^3 = 343 < 493 < 8^3 = 512$$


The smaller cube is $7^3 = 343$, so the ten's digit is $7$.

Step 4: Combine digits:

$$\mathbf{\sqrt[3]{493039} = 79}$$

.


Group 1 ($039$) ends in 9 $\to$ unit digit 9. Group 2 ($493$) lies between $7^3$ (343) and $8^3$ (512) $\to$ tens digit 7. Root is 79.
5
Evaluate: $\sqrt[3]{\frac{729}{2197}}$.
उत्तर एवं व्याख्या देखें
उत्तर: Apply the Quotient Rule $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$:
• Find $\sqrt[3]{729}$:
$$729 = 9 \times 9 \times 9 = 3^6 \implies \sqrt[3]{729} = 9$$
• Find $\sqrt[3]{2197}$:
$$2197 = 13 \times 13 \times 13 \implies \sqrt[3]{2197} = 13$$
$$\mathbf{\sqrt[3]{\frac{729}{2197}} = \frac{9}{13}}$$.
$\sqrt[3]{729} = 9$ and $\sqrt[3]{2197} = 13$. Result is $9/13$.
6
Evaluate: $\sqrt[3]{0.000125}$.
उत्तर एवं व्याख्या देखें
उत्तर: Convert the decimal into a fraction:
$$0.000125 = \frac{125}{1,000,000}$$
Apply the cube root quotient rule:
$$\sqrt[3]{\frac{125}{1,000,000}} = \frac{\sqrt[3]{125}}{\sqrt[3]{1,000,000}}$$
• $\sqrt[3]{125} = 5$ (since $5^3 = 125$)
• $\sqrt[3]{1,000,000} = 100$ (since $100^3 = 1,000,000$)
$$= \frac{5}{100} = \mathbf{0.05}$$.
Convert to $125 / 1,000,000$. Cube root is $5 / 100 = 0.05$.
7
Three numbers are in the ratio $1 : 2 : 3$ and the sum of their cubes is $4500$. Find the three numbers.
उत्तर एवं व्याख्या देखें
उत्तर: Let the three numbers be $x, 2x,$ and $3x$.
According to the problem:
$$x^3 + (2x)^3 + (3x)^3 = 4500$$
$$x^3 + 8x^3 + 27x^3 = 4500$$
$$36x^3 = 4500$$
Divide by $36$:
$$x^3 = \frac{4500}{36} = 125$$
Take the cube root of both sides:
$$x = \sqrt[3]{125} = 5$$
• The three numbers are:
1st number $= x = \mathbf{5}$
2nd number $= 2x = 2(5) = \mathbf{10}$
3rd number $= 3x = 3(5) = \mathbf{15}$.
*(Check: $5^3 + 10^3 + 15^3 = 125 + 1000 + 3375 = 4500$)*.
$x^3 + 8x^3 + 27x^3 = 36x^3 = 4500 \implies x^3 = 125 \implies x = 5$. Numbers are 5, 10, 15.
8
The volume of a cubical water reservoir is $91.125\text{ cubic meters}$. Find the length of its edge in meters.
उत्तर एवं व्याख्या देखें
उत्तर: Volume of cube $V = s^3 = 91.125\text{ m}^3$.
$$\text{Side } s = \sqrt[3]{91.125} = \sqrt[3]{\frac{91125}{1000}}$$
• Find $\sqrt[3]{91125}$ by prime factorisation or estimation:
Group 1: $125 \to 5$; Group 2: $91 \to 4^3 = 64 < 91 < 5^3 = 125 \to 4$. So $\sqrt[3]{91125} = 45$.
• Find $\sqrt[3]{1000} = 10$.
$$\text{Side } s = \frac{45}{10} = \mathbf{4.5\text{ meters}}$$.
$\sqrt[3]{91.125} = \sqrt[3]{91125 / 1000} = 45 / 10 = 4.5\text{ meters}$.
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कक्षा 8 Mathematics के सभी अध्याय

अध्याय 1: गुणनखंडन (Factorisation) अध्याय 2: परिमेय संख्याएँ (Rational Numbers) अध्याय 3: घातांक (Exponents) अध्याय 4: वर्ग और वर्गमूल (Squares and Square Roots) अध्याय 5: घन और घनमूल (Cubes and Cube Roots) अध्याय 6: संख्याओं के साथ खेलना (Playing With Numbers) अध्याय 7: समुच्चय (Sets) अध्याय 8: प्रतिशतता (Percentage) अध्याय 9: लाभ, हानि और छूट (Profit, Loss and Discount) अध्याय 10: सरल एवं चक्रवृद्धि ब्याज (Simple and Compound Interest) अध्याय 11: प्रत्यक्ष और व्युत्क्रम भिन्नताएँ (Direct and Inverse Variations) अध्याय 12: बीजीय व्यंजक (Algebraic Expressions) अध्याय 13: बीजगणितीय सर्वसमिकाएँ (Algebraic Identities) अध्याय 14: रैखिक असमिकाएँ (Linear Inequations) अध्याय 15: रैखिक समीकरण (Linear Equations) अध्याय 16: आकारों की समझ (Understanding Shapes) अध्याय 17: त्रिविमीय आकारों का द्विविमीय निरूपण (Representing 3D in 2D) अध्याय 18: विशेष प्रकार के चतुर्भुजों का निर्माण (Construction of Special Types of Quadrilaterals) अध्याय 19: आलेखों से परिचय (Introduction to Graphs) अध्याय 20: रेखा समरूपता और परावर्तन (Line Symmetry and Reflection) अध्याय 21: घेरा (Circle) अध्याय 22: सरलरेखीय आकृतियों का क्षेत्रफल (Area of Rectilinear Figures) अध्याय 23: ठोसों का सतही क्षेत्रफल और आयतन (Surface Area and Volume of Solids) अध्याय 24: आँकड़ों का प्रबंधन (Data Handling) अध्याय 25: प्रायिकता (Probability)

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