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Cubes and Cube Roots - Cubes and Perfect Cubes

Grade 8CBSE

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

🔑Concepts

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A cube of a number is the result of multiplying that number by itself three times. For a number nn, its cube is represented as n3=n×n×nn^3 = n \times n \times n. Visually, this can be imagined as a 3D solid cube where the length, width, and height are all equal to nn units, and the total number of unit blocks making up the cube represents the cube value.

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A natural number is called a perfect cube if it is the cube of some natural number. For example, 6464 is a perfect cube because 64=4×4×464 = 4 \times 4 \times 4. If you visualize a set of 6464 small blocks, they can be perfectly arranged to form one large solid cube with 44 blocks along each edge.

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The units digit of the cube of a number depends only on the units digit of the number itself. For numbers ending in 1,4,5,6,9,1, 4, 5, 6, 9, or 00, their cubes also end 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 numbers are always even, and cubes of odd numbers are always odd. For example, 23=82^3 = 8 (even) and 33=273^3 = 27 (odd).

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Prime Factorization Rule: A number is a perfect cube only if each prime factor in its prime factorization appears in a group of three (triplets). If you visualize the factors written out, you should be able to circle them in sets of three identical numbers with none left over.

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Interesting Pattern with Odd Numbers: Cubes can be expressed as the sum of consecutive odd numbers. For example, 13=11^3 = 1; 23=3+5=82^3 = 3 + 5 = 8; 33=7+9+11=273^3 = 7 + 9 + 11 = 27. This forms a visual ladder where each subsequent cube requires the next set of consecutive odd numbers.

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The cube of a negative number is always negative. Since (−a)3=(−a)×(−a)×(−a)(-a)^3 = (-a) \times (-a) \times (-a), the first two negatives become positive, and the final multiplication by a negative result makes the entire product negative, such as (−2)3=−8(-2)^3 = -8.

📐Formulae

Volume of a cube with side ss: V=s3V = s^3

Cube of a product: (a×b)3=a3×b3(a \times b)^3 = a^3 \times b^3

Cube of a fraction: (ab)3=a3b3(\frac{a}{b})^3 = \frac{a^3}{b^3}

Negative base: (−a)3=−a3(-a)^3 = -a^3

General form of a perfect cube: n=m3 where m∈Nn = m^3 \text{ where } m \in \mathbb{N}

💡Examples

Problem 1:

Check if 216 is a perfect cube using prime factorization.

Solution:

Step 1: Perform prime factorization of 216. 216=2×108216 = 2 \times 108 216=2×2×54216 = 2 \times 2 \times 54 216=2×2×2×27216 = 2 \times 2 \times 2 \times 27 216=2×2×2×3×9216 = 2 \times 2 \times 2 \times 3 \times 9 216=2×2×2×3×3×3216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 Step 2: Group the prime factors into triplets. 216=(2×2×2)×(3×3×3)216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) Step 3: Since all prime factors occur in triplets, 216 is a perfect cube. 216=23×33=(2×3)3=63216 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3

Explanation:

To determine if a number is a perfect cube, we break it down into its prime factors and check if every factor can be grouped into sets of three. Since both 2 and 3 form complete triplets, the number is a perfect cube.

Problem 2:

Find the smallest number by which 72 must be multiplied so that the product is a perfect cube.

Solution:

Step 1: Prime factorize 72. 72=2×2×2×3×372 = 2 \times 2 \times 2 \times 3 \times 3 Step 2: Group the factors into triplets. 72=(2×2×2)×3×372 = (2 \times 2 \times 2) \times 3 \times 3 Step 3: Observe the factors that do not form a triplet. The prime factor 3 appears only twice (3×33 \times 3). Step 4: To make it a triplet, we need one more 3. Therefore, the smallest number to multiply is 3. Check: 72×3=21672 \times 3 = 216, which is 636^3.

Explanation:

By examining the prime factorization, we identify which factors are missing to complete a group of three. Here, we have a triplet of 2s but only two 3s, so multiplying by one more 3 completes the pattern.