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A Square and A Cube - Cubic Numbers

Grade 8CBSE

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

🔑Concepts

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A cube number is obtained when a number is multiplied by itself three times. For any number nn, its cube is n3=n×n×nn^3 = n \times n \times n.

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A number is called a perfect cube if it is the cube of some natural number. For example, 216216 is a perfect cube because 216=63216 = 6^3.

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Cubes of even numbers are always even (e.g., 23=82^3 = 8, 43=644^3 = 64), and cubes of odd numbers are always odd (e.g., 33=273^3 = 27, 53=1255^3 = 125).

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Properties of unit digits: If a number ends in 0,1,4,5,6,90, 1, 4, 5, 6, 9, its cube also ends in the same digit. If a number ends in 22, its cube ends in 88 (and vice versa). If a number ends in 33, its cube ends in 77 (and vice versa).

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The cube root of a number xx is denoted by x3\sqrt[3]{x}. It is the value that, when cubed, gives xx. For example, 1253=5\sqrt[3]{125} = 5.

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To check if a number is a perfect cube using Prime Factorization, every prime factor must appear in groups of three (triplets).

📐Formulae

n3=n×n×nn^3 = n \times n \times n

x3=y  ⟺  y3=x\sqrt[3]{x} = y \iff y^3 = x

(a×b)3=a3×b3(a \times b)^3 = a^3 \times b^3

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

💡Examples

Problem 1:

Is 243243 a perfect cube? If not, find the smallest number by which 243243 must be multiplied to make it a perfect cube.

Solution:

Prime factorization of 243243 is: 243=3×3×3×3×3=33×32243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^3 \times 3^2. Since the second group of 33 is not a triplet (it only has two 33s), 243243 is not a perfect cube. To make it a perfect cube, we need one more 33. 243×3=729243 \times 3 = 729, and 729=93729 = 9^3.

Explanation:

We group the prime factors in triplets. Any factor that does not form a complete triplet indicates the number is not a perfect cube.

Problem 2:

Find the cube root of 80008000 using prime factorization.

Solution:

8000=2×2×2×2×2×2×5×5×58000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 Grouping them into triplets: 8000=(2×2×2)×(2×2×2)×(5×5×5)8000 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) 8000=23×23×538000 = 2^3 \times 2^3 \times 5^3 80003=2×2×5=20\sqrt[3]{8000} = 2 \times 2 \times 5 = 20

Explanation:

By expressing the number as a product of its prime factors and grouping them into threes, we can determine the cube root by taking one factor from each triplet.

Problem 3:

Find the smallest number by which 128128 must be divided to obtain a perfect cube.

Solution:

Prime factorization of 128128: 128=2×2×2×2×2×2×2128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 128=(2×2×2)×(2×2×2)×2128 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times 2 128=23×23×2128 = 2^3 \times 2^3 \times 2. The factor 22 does not form a triplet. Therefore, we must divide 128128 by 22: 128÷264\begin{array}{r} 128 \div 2 \\ \hline 64 \end{array} 64=4364 = 4^3, which is a perfect cube.

Explanation:

To make a number a perfect cube by division, we remove the prime factors that do not form complete triplets.