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Cubes and Cube Roots - Properties of Cube Numbers

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of a Cube: The cube of a number is the product obtained by multiplying the number by itself three times. For any number nn, its cube is denoted as n3=n×n×nn^3 = n \times n \times n. Visually, if you have a solid 3D cube with a side length of nn units, the total volume or the number of unit cubes contained within it represents n3n^3.

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Units Digit of Cubes: The units digit of the cube of a number is determined solely by the units digit of the number itself. Numbers ending in 0,1,4,5,6,90, 1, 4, 5, 6, 9 have cubes ending in the same digit. However, if a number ends in 22, its cube ends in 88 (and vice versa); if it ends in 33, its cube ends in 77 (and vice versa).

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Cubes of Even and Odd Numbers: The cube of every even number is even (e.g., 23=82^3 = 8, 43=644^3 = 64), and the cube of every odd number is odd (e.g., 33=273^3 = 27, 53=1255^3 = 125).

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Cubes of Negative Numbers: The cube of a negative number is always negative. For example, (−5)3=(−5)×(−5)×(−5)=−125(-5)^3 = (-5) \times (-5) \times (-5) = -125. This differs from squaring, where the result is always positive.

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Perfect Cubes and Prime Factorization: A natural number is called a perfect cube if it can be expressed as the product of triplets of equal prime factors. For instance, 216=2×2×2×3×3×3216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3. Visually, you can imagine these factors grouped into sets of three identical blocks; if every block has two partners to form a trio, the number is a perfect cube.

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Adding Consecutive Odd Numbers: Cubes can be represented as the sum of consecutive odd numbers. 13=11^3 = 1, 23=3+5=82^3 = 3 + 5 = 8, 33=7+9+11=273^3 = 7 + 9 + 11 = 27, and so on. To find the sum for n3n^3, you start with the odd number following the last one used for (n−1)3(n-1)^3.

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Cube Roots: The cube root is the inverse operation of finding a cube. If x3=yx^3 = y, then the cube root of yy, denoted by y3\sqrt[3]{y}, is xx. Geometrically, if you are given the volume of a cube, the cube root represents the length of one of its edges.

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Cubes of Rational Numbers: The cube of a fraction ab\frac{a}{b} is calculated by cubing the numerator and the denominator separately, resulting in a3b3\frac{a^3}{b^3}.

📐Formulae

n3=n×n×nn^3 = n \times n \times n

x3=x13\sqrt[3]{x} = x^{\frac{1}{3}}

(ab)3=a3×b3(ab)^3 = a^3 \times b^3

(ab)3=a3b3(\frac{a}{b})^3 = \frac{a^3}{b^3}

ab3=a3×b3\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}

ab3=a3b3\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}

Sum of first nn odd numbers: 1+3+5+...+(2n−1)=n21 + 3 + 5 + ... + (2n-1) = n^2 (Note: This is for squares, but relevant for understanding number patterns in cubes)

💡Examples

Problem 1:

Determine if 13311331 is a perfect cube using prime factorization.

Solution:

Step 1: Perform prime factorization of 13311331. We find that 1331=11×121=11×11×111331 = 11 \times 121 = 11 \times 11 \times 11. \ Step 2: Group the prime factors into triplets. The factors are (11×11×11)(11 \times 11 \times 11). \ Step 3: Since there are no remaining factors outside the triplet, 13311331 is a perfect cube. \ Thus, 13313=11\sqrt[3]{1331} = 11.

Explanation:

To check for a perfect cube, we decompose the number into its prime factors and check if every prime factor appears in a group of three.

Problem 2:

Find the smallest number by which 675675 must be multiplied so that the product is a perfect cube.

Solution:

Step 1: Prime factorize 675675. \ 675=5×135=5×5×27=5×5×3×3×3675 = 5 \times 135 = 5 \times 5 \times 27 = 5 \times 5 \times 3 \times 3 \times 3. \ Step 2: Group the factors into triplets: (3×3×3)×(5×5)(3 \times 3 \times 3) \times (5 \times 5). \ Step 3: Identify the incomplete triplet. The factor 33 forms a complete triplet, but the factor 55 only appears twice. \ Step 4: To make it a triplet, we need one more 55. \ Therefore, the smallest number to multiply by is 55. \ New number: 675×5=3375675 \times 5 = 3375, which is 15315^3.

Explanation:

We use prime factorization to identify which factor does not form a complete group of three. Multiplying by the missing factors completes the triplet to form a perfect cube.