Cubes and Cube Roots

Cube Numbers and Perfect Cubes: Definition and Properties

Understanding Cubing a Number

The operation of cubing a number means multiplying the number by itself three times. For example, cubing the number 4 is $4 \times 4 \times 4 = 64$. This operation is represented using an exponent of 3. $4^3 = 64$. When an integer is cubed, the resulting number is called a cube number or a perfect cube.

Definition of Cube Numbers and Perfect Cubes

A cube number, also known as a perfect cube, is an integer $n$ that is the cube of another integer $m$. That is, $n = m \times m \times m$, or in exponential notation, $n = m^3$, for some integer $m$.

We commonly refer to the cubes of whole numbers ($m \in \{0, 1, 2, 3, \ldots\}$) when listing non-negative perfect cubes.

Examples of perfect cubes (by cubing the first few integers):

• $0^3 = 0 \times 0 \times 0 = 0$. So, 0 is a perfect cube.
• $1^3 = 1 \times 1 \times 1 = 1$. So, 1 is a perfect cube.
• $2^3 = 2 \times 2 \times 2 = 8$. So, 8 is a perfect cube.
• $3^3 = 3 \times 3 \times 3 = 27$. So, 27 is a perfect cube.
• $4^3 = 4 \times 4 \times 4 = 64$. So, 64 is a perfect cube.
• $5^3 = 5 \times 5 \times 5 = 125$. So, 125 is a perfect cube.
• $10^3 = 10 \times 10 \times 10 = 1000$. So, 1000 is a perfect cube.

The sequence of the first few non-negative perfect cubes is $\mathbf{0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, ...}$

Unlike squaring, cubing a negative integer results in a negative number:

• $(-1)^3 = (-1) \times (-1) \times (-1) = 1 \times (-1) = -1$. So, -1 is a perfect cube.
• (-2)^3 = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$. So, -8 is a perfect cube.
• (-3)^3 = (-3) \times (-3) \times (-3) = 9 \times (-3) = -27$. So, -27 is a perfect cube.

Perfect cubes can be positive, negative, or zero. The sign of a perfect cube is the same as the sign of the integer that was cubed.

Properties of Perfect Cubes

Perfect cubes exhibit several properties that can help in their identification and understanding:

1. Ending Digits (Unit's Place):

   The last digit (the digit in the units place) of a perfect cube can be *any* digit from $0$ to $9$. Unlike perfect squares, which have restricted ending digits, the last digit of a perfect cube can be any of the ten digits.

   Let's observe the last digit of the cube of each digit from 0 to 9:

   $$ 0^3 = 0 \quad (\text{ends in } 0) $$ $$ 1^3 = 1 \quad (\text{ends in } 1) $$ $$ 2^3 = 8 \quad (\text{ends in } 8) $$ $$ 3^3 = 27 \quad (\text{ends in } 7) $$ $$ 4^3 = 64 \quad (\text{ends in } 4) $$ $$ 5^3 = 125 \quad (\text{ends in } 5) $$ $$ 6^3 = 216 \quad (\text{ends in } 6) $$ $$ 7^3 = 343 \quad (\text{ends in } 3) $$ $$ 8^3 = 512 \quad (\text{ends in } 2) $$ $$ 9^3 = 729 \quad (\text{ends in } 9) $$

   All digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ appear as the last digit of a perfect cube. This means that the last digit of a number alone does not tell you if it is a perfect cube (except for confirming it's *possibly* a cube). Note that some endings uniquely determine the base's ending digit: e.g., a cube ending in 8 implies the base ended in 2; ending in 7 implies the base ended in 3; ending in 3 implies base ended in 7; ending in 2 implies base ended in 8. Other endings (0, 1, 4, 5, 6, 9) are not unique.

2. Number of Zeros at the End:

   If a perfect cube ends with one or more zeros, the number of trailing zeros must always be a multiple of three (i.e., $3k$ for some integer $k \ge 1$).

   Example: $1000 = 10^3$ ends in 3 zeros. $8000 = 20^3 = (2 \times 10)^3 = 2^3 \times 10^3 = 8 \times 1000$ ends in 3 zeros. $27000000 = 300^3 = (3 \times 100)^3 = 3^3 \times (10^2)^3 = 3^3 \times 10^6 = 27 \times 1000000$ ends in 6 zeros.

   Explanation: A trailing zero indicates a factor of 10. If a positive integer $m$ ends in exactly $k$ zeros, its prime factorization contains $2^k$ and $5^k$ (or higher powers of 2 or 5, but at least $k$ factors of each). When we cube $m$, $m^3 = (p_1^{e_1} \ldots 2^k 5^k \ldots)^3 = p_1^{3e_1} \ldots 2^{3k} 5^{3k} \ldots$. The number of trailing zeros in $m^3$ is determined by the minimum of the powers of 2 and 5 in its prime factorization, which will be $3k$ if $m$ didn't have extra factors of 2 or 5 beyond those forming the trailing zeros. If $m$ has $j$ trailing zeros, $m$ is a multiple of $10^j$, and $m^3$ is a multiple of $(10^j)^3 = 10^{3j}$, so $m^3$ ends in at least $3j$ zeros. For a perfect cube, the total number of trailing zeros is always a multiple of 3.

3. Prime Factorization:

   In the unique prime factorization of a perfect cube, the exponent of every prime factor is always a multiple of three.

   Explanation: If an integer $n$ is a perfect cube, then $n = m^3$ for some integer $m$. Let the unique prime factorization of $m$ be $m = \pm p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}$ (for integers, the $\pm$ sign handles negative bases, and 0 is $0^3$). Then the prime factorization of $n$ is obtained by cubing the factorization of $m$: $n = (\pm p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k})^3 = (\pm 1)^3 p_1^{3e_1} p_2^{3e_2} \ldots p_k^{3e_k} = (\pm 1) p_1^{3e_1} p_2^{3e_2} \ldots p_k^{3e_k}$. Each exponent in the factorization of $n$ (ignoring the sign for negative $n$) is of the form $3e_i$, which is a multiple of 3.

   Conversely, if the prime factorization of a number $n$ (ignoring a possible leading negative sign if $n<0$) has all exponents as multiples of three, $n = p_1^{3e_1} p_2^{3e_2} \ldots p_k^{3e_k}$. This can be rewritten as $n = (p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k})^3$. Since $e_i$ are integers, $m = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}$ is also an integer (or its negative), and $n = m^3$ is a perfect cube.

   Example: $64$. Prime factorization: $64 = 2^6$. The exponent is 6, which is a multiple of 3 ($6 = 3 \times 2$). $64 = (2^2)^3 = 4^3$.

   Example: $216$. Prime factorization: $216 = 2^3 \times 3^3$. The exponents are 3 and 3 (both multiples of 3). $216 = (2^1 \times 3^1)^3 = 6^3$.

   Example: $1728$. Prime factorization: $1728 = 2^6 \times 3^3$. Exponents are 6 and 3 (both multiples of 3). $1728 = (2^2 \times 3^1)^3 = (4 \times 3)^3 = 12^3$.

   Example: $12 = 2^2 \times 3^1$. The exponent of 2 is 2 (not a multiple of 3). The exponent of 3 is 1 (not a multiple of 3). So 12 is not a perfect cube.

4. Between Consecutive Cubes:

   Between the cubes of two consecutive integers, $n^3$ and $(n+1)^3$, there are exactly $(n+1)^3 - n^3 - 1$ non-perfect cube integers.

   The number of integers is $(n^3 + 3n^2 + 3n + 1) - n^3 - 1 = 3n^2 + 3n$. None of these integers are perfect cubes because if $k^3$ was one of them, then $n^3 < k^3 < (n+1)^3$. Taking the cube root would give $n < k < n+1$. But there is no integer $k$ strictly between two consecutive integers $n$ and $n+1$ (for $n \in \mathbb{Z}$). Thus, the integers between $n^3$ and $(n+1)^3$ are all non-perfect cubes.

   Example: Between $2^3 = 8$ and $3^3 = 27$. The number of non-perfect cubes is $3(2^2) + 3(2) = 3(4) + 6 = 12 + 6 = 18$. These are $9, 10, ..., 26$. The count is $26 - 9 + 1 = 18$.

These properties, particularly the prime factorization rule, are the most reliable ways to determine if a number is a perfect cube and are essential for finding cube roots.

Cubes of Negative Numbers and Rational Numbers

The operation of cubing (raising to the power of 3) can be applied to any real number, including integers (positive, negative, and zero), fractions, decimals, and irrational numbers. The process remains the same: multiply the number by itself three times ($m^3 = m \times m \times m$). The result of cubing an integer is a perfect cube, as discussed in the previous section.

Cubes of Negative Numbers

When a negative number is cubed, the result is always a negative number. This can be understood by considering the sign rules for multiplication:

A negative number multiplied by a negative number gives a positive number ($(-) \times (-) = (+)$).

Multiplying this positive result by another negative number gives a negative result ($ (+) \times (-) = (-)$).

So, $(-a)^3 = (-a) \times (-a) \times (-a)$.

This shows that the cube of a negative number is the negative of the cube of its absolute value.

Examples:

• $(-1)^3 = (-1) \times (-1) \times (-1) = 1 \times (-1) = -1$.
• $(-2)^3 = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$.
• $(-5)^3 = (-5) \times (-5) \times (-5) = 25 \times (-5) = -125$.
• $(-10)^3 = (-10) \times (-10) \times (-10) = 100 \times (-10) = -1000$.

Perfect cubes can therefore be positive (the cubes of positive integers), negative (the cubes of negative integers), or zero (the cube of zero). The sign of a perfect cube matches the sign of the integer that was cubed.

Cubes of Rational Numbers (Fractions and Decimals)

Cubing a rational number, whether it's expressed as a fraction $\frac{p}{q}$ or as a terminating or repeating decimal, follows the definition of repeated multiplication. Using the laws of exponents, specifically the power of a quotient law $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$, cubing a fraction $\frac{p}{q}$ means cubing the numerator and cubing the denominator.

Examples of cubes of rational numbers (fractions):

• $(\frac{1}{2})^3 = \frac{1^3}{2^3} = \frac{1 \times 1 \times 1}{2 \times 2 \times 2} = \frac{1}{8}$.
• $(\frac{2}{3})^3 = \frac{2^3}{3^3} = \frac{8}{27}$.
• $(\frac{-1}{4})^3 = \frac{(-1)^3}{4^3} = \frac{-1}{64} = -\frac{1}{64}$.
• $(\frac{-2}{5})^3 = \frac{(-2)^3}{5^3} = \frac{-8}{125} = -\frac{8}{125}$.

For decimal numbers, you can either convert them to fractions and cube the fraction, or multiply the decimal by itself three times.

Examples of cubes of rational numbers (decimals):

• $(0.1)^3 = 0.1 \times 0.1 \times 0.1 = 0.01 \times 0.1 = 0.001$. (Number of decimal places triples when cubing).
• $(0.2)^3 = 0.2 \times 0.2 \times 0.2 = 0.04 \times 0.2 = 0.008$.
• $(1.1)^3 = 1.1 \times 1.1 \times 1.1 = 1.21 \times 1.1 = 1.331$.
• $(-0.5)^3 = (-0.5) \times (-0.5) \times (-0.5) = 0.25 \times (-0.5) = -0.125$.

A rational number $\frac{p}{q}$ is a perfect cube if and only if both the numerator $p$ and the denominator $q$ are perfect cubes (considering the sign). For example, $\frac{8}{27} = \frac{2^3}{3^3}$, both 8 and 27 are perfect cubes, so $\frac{8}{27}$ is a perfect cube $(\frac{2}{3})^3$. However, $\frac{4}{9}$ is not a perfect cube because 4 and 9 are not perfect cubes, even though they are perfect squares.

This understanding of cubing negative and rational numbers sets the stage for finding the inverse operation, the cube root, for these types of numbers.

Cube Roots: Definition and Methods (Prime Factorisation, Estimation)

Definition of Cube Root

Finding the cube root of a number is the inverse operation of cubing a number. If cubing a number $m$ results in $x$ (i.e., $m^3 = x$), then the cube root of $x$ is $m$.

The cube root of a number $x$ is denoted by the radical symbol $\sqrt{\phantom{x}}$ with a small index $3$ placed in the hook: $\sqrt[3]{x}$. The index $3$ indicates that it is the cube root.

Unlike square roots of positive numbers (which have two real roots), every real number (positive, negative, or zero) has exactly one real cube root.

• $\sqrt[3]{8} = 2$, because $2 \times 2 \times 2 = 2^3 = 8$. (The cube root of a positive number is positive).
• $\sqrt[3]{-8} = -2$, because $(-2) \times (-2) \times (-2) = (-2)^3 = -8$. (The cube root of a negative number is negative).
• $\sqrt[3]{0} = 0$, because $0 \times 0 \times 0 = 0^3 = 0$.
• $\sqrt[3]{1} = 1$, because $1^3 = 1$.
• $\sqrt[3]{-1} = -1$, because $(-1)^3 = -1$.
• $\sqrt[3]{64} = 4$, because $4^3 = 64$.
• $\sqrt[3]{-125} = -5$, because $(-5)^3 = -125$.

If an integer is a perfect cube, its cube root is an integer. If a real number is not a perfect cube, its real cube root is an irrational number (unless the number is 1, in which case the cube root is 1). For example, $\sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{9}, \sqrt[3]{10}$ are irrational numbers.

Methods for Finding Cube Roots

There are several methods to find the cube root of a number. For perfect cubes, prime factorization is efficient. For numbers that are not perfect cubes, estimation or numerical methods (like the long division method for cube roots, which is more complex than for square roots) are used.

Method 1: Prime Factorisation Method (Suitable for perfect cubes)

This is the primary method for finding the cube root of a perfect cube. It is based on the property that in the unique prime factorization of a perfect cube, the exponent of every prime factor is a multiple of three. Finding the cube root involves dividing each exponent by 3.

Steps:

1. Find the prime factorization of the given number. Express the factorization in exponential form. If the number is negative, ignore the sign for now and find the prime factorization of its absolute value.
2. Examine the exponents of all the prime factors. If all the exponents are multiples of three, the absolute value of the number is a perfect cube. If any exponent is not a multiple of three, the absolute value of the number is not a perfect cube.
3. To find the cube root, divide each of the exponents in the prime factorization by $3$.
4. Multiply the prime factors raised to these new exponents. This gives the magnitude of the cube root.
5. Determine the sign of the cube root: If the original number was positive, the cube root is positive. If the original number was negative, the cube root is negative. The cube root of 0 is 0.

Answer:

Answer:

If a number is not a perfect cube, the prime factorization method will show that some prime factors have exponents that are not multiples of three, indicating that its real cube root is an irrational number (e.g., $\sqrt[3]{4} = \sqrt[3]{2^2}$, the exponent 2 is not a multiple of 3). This method can still be used to simplify the radical, if possible (e.g., $\sqrt[3]{16} = \sqrt[3]{2^4} = \sqrt[3]{2^3 \times 2^1} = \sqrt[3]{2^3} \times \sqrt[3]{2^1} = 2\sqrt[3]{2}$).

Method 2: Estimation Method

For numbers that are not perfect cubes, we can estimate their cube roots by finding the two consecutive perfect cubes between which the given number lies. This gives us the integer part of the cube root.

Steps to estimate $\sqrt[3]{N}$ for a real number $N$:

1. Find the two consecutive integers $m$ and $m+1$ such that $m^3 \le N < (m+1)^3$. (If $N$ is negative, find integers $m$ and $m+1$ such that $(m+1)^3 < N \le m^3$. For example, $-27 < -10 < -8$, so $\sqrt[3]{-27} < \sqrt[3]{-10} < \sqrt[3]{-8}$, i.e., $-3 < \sqrt[3]{-10} < -2$).
2. Then, the real cube root $\sqrt[3]{N}$ is a number between $m$ and $m+1$.
3. To get a slightly better estimate, consider whether $N$ is closer to $m^3$ or $(m+1)^3$. If $N$ is closer to $m^3$, $\sqrt[3]{N}$ will be closer to $m$. If $N$ is closer to $(m+1)^3$, $\sqrt[3]{N}$ will be closer to $m+1$.

Answer:

For more precise approximations of cube roots of non-perfect cubes, numerical methods (like iterative methods or using logarithms and antilogarithms) or calculators are typically used.

Cube Roots of Negative Numbers, Products, and Rational Numbers

The operation of finding the cube root can be applied to any real number. Unlike square roots of positive numbers, which yield two real roots, cube roots always yield a single real root for any real number. Understanding the cube root of negative numbers and applying rules for products and quotients is important for simplifying expressions involving cube roots.

Cube Roots of Negative Numbers

As discussed in the previous section, cubing a negative number results in a negative number (e.g., $(-2)^3 = -8$). Conversely, the cube root of a negative number is always a negative real number. Every negative real number has exactly one real cube root, and that root is negative.

The cube root of a negative number $-x$ (where $x$ is a positive real number) is the negative of the cube root of the positive number $x$.

Example: Find $\sqrt[3]{-27}$.

We find the cube root of the positive number 27: $\sqrt[3]{27} = 3$ (since $3^3 = 27$).

Since the original number is negative (-27), its cube root is the negative of $\sqrt[3]{27}$.

Check the result: $(-3)^3 = (-3) \times (-3) \times (-3) = 9 \times (-3) = -27$. Correct.

This property confirms that every real number (positive, negative, or zero) has exactly one real cube root.

Cube Root of a Product

The cube root of a product of two or more numbers is equal to the product of their cube roots. This property holds for any real numbers $a$ and $b$.

This property is particularly useful for simplifying cube roots by factoring out perfect cube factors from the number under the radical sign.

Example: Simplify $\sqrt[3]{32}$.

Find factors of 32, specifically looking for perfect cube factors. $32 = 8 \times 4$. 8 is a perfect cube ($2^3$).

Using the property of the cube root of a product:

Evaluate the cube root of the perfect cube factor $\sqrt[3]{8}=2$. Since 4 is not a perfect cube, $\sqrt[3]{4}$ is irrational and cannot be simplified further using integer factors.

So, $\sqrt[3]{32} = \mathbf{2\sqrt[3]{4}}$.

Cube Root of a Quotient

The cube root of a quotient (a fraction) of two numbers is equal to the quotient of their cube roots, provided the denominator is not zero.

This property is useful for finding the cube root of fractions and is also used in the process of rationalizing denominators involving cube roots.

Example: Find $\sqrt[3]{\frac{64}{125}}$.

Using the property of the cube root of a quotient:

Evaluate the cube roots of the numerator and denominator. We know $4^3 = 64$ and $5^3 = 125$.

Check: $(\frac{4}{5})^3 = \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} = \frac{4^3}{5^3} = \frac{64}{125}$. Correct.

So, $\sqrt[3]{\frac{64}{125}} = \mathbf{\frac{4}{5}}$.

Cube Roots of Rational Numbers (Fractions and Decimals)

A rational number is a perfect cube if and only if its cube root is also a rational number. For a fraction $\frac{p}{q}$ (in simplest form), it is a perfect cube if and only if its numerator $p$ and its denominator $q$ are both perfect cubes (considering the sign). The cube root is found by taking the cube root of the numerator and the cube root of the denominator.

For decimal numbers, you can find the cube root by converting the decimal to a fraction and then applying the rule for fractions, or by using estimation/approximation methods. If a decimal number is a perfect cube, its decimal representation must have a number of decimal places that is a multiple of three. The number of decimal places in the cube root will be one-third of the number of decimal places in the original decimal.

Example: Find $\sqrt[3]{0.008}$.

The number of decimal places in 0.008 is 3, which is a multiple of three. Remove the decimal point and find the cube root of the resulting integer: $\sqrt[3]{8} = 2$. The cube root will have $3 \div 3 = 1$ decimal place. Place the decimal point in 2 to get 1 decimal place: 0.2.

Check: $(0.2)^3 = 0.2 \times 0.2 \times 0.2 = 0.04 \times 0.2 = 0.008$. Correct.

If a rational number is not a perfect cube, its real cube root will be an irrational number. For instance, $\sqrt[3]{\frac{1}{2}} = \frac{\sqrt[3]{1}}{\sqrt[3]{2}} = \frac{1}{\sqrt[3]{2}}$. The denominator $\sqrt[3]{2}$ is irrational. To rationalize this denominator, we need to make the number under the cube root a perfect cube. Multiply the numerator and denominator by $\sqrt[3]{2^2} = \sqrt[3]{4}$:

$\frac{1}{\sqrt[3]{2}} \times \frac{\sqrt[3]{4}}{\sqrt[3]{4}} = \frac{\sqrt[3]{1 \times 4}}{\sqrt[3]{2 \times 4}} = \frac{\sqrt[3]{4}}{\sqrt[3]{8}} = \frac{\sqrt[3]{4}}{2}$.

The concepts of cubes and cube roots, along with their properties for negative numbers and rational numbers, are essential for working with powers and radicals with an index of 3.