Cube and Cube Roots

Ratio and Proportion

The relation between two quantities (both of the same kind and in the same unit) obtained on dividing one quantity by the other, is called the ratio.

Algebra

An expression made up of variables (or literals) and constants (or numerals) connected by signs of arithmetic operations is called an algebraic expression.

Understanding Elementary Shapes

When we measure a line segment, we measure its length or distance from one end point to the other.

Decimals

Decimals are another way of writing part of a whole number. In a place value table on moving a digit from the right to the left by one place...

Whole Numbers

The number which comes immediately after a particular number is called its successor.

Cubes And Cube Roots

1.0Cubes

If we multiply a number by itself three times, then it is the cube of that number. Cubes in geometry, is a solid figure with 6 faces (which are congruent squares), with 12 edges (which are equal in length) and with 8 vertices. If a cube has an edge of 1 cm , its volume $= 1 \times 1 \times 1 = 1 cm^{3}$ . Similarly, if the edge is 2 cm , its volume $= 2 \times 2 \times 2 = 8 cm^{3}$ . Here 1, 8 are perfect cubes. But 9 is not a perfect cube because there is no natural number which multiplied by itself three times gives 9 . The following table gives the cubes of the first 20 natural numbers.

Number	Cube	Number	Cube
1	1	11	1331
2	8	12	1728
3	27	13	2197
4	64	14	2744
5	125	15	3375
6	216	16	4096
7	343	17	4913
8	512	18	5832
9	729	19	6859
10	1000	20	8000

Some important points

(i) Cube of a rational number $\frac{a}{b}$ is $\frac{a ^{3}}{b ^{3}}$ i.e., $(\frac{a}{b})^{3} = \frac{a ^{3}}{b ^{3}}$ . (ii) Cube of the product of number is equal to the product of their cubes. i.e. $(a \times b)^{3} = a^{3} \times b^{3}$ .

Q. Evaluate: (i) $[4 \times (- 5)]^{3}$ (ii) $(- \frac{4}{5})^{3}$ Explanation: (i) $[4 \times (- 5)]^{3} = 4^{3} \times (- 5)^{3} = 64 \times (- 125) = - 8000$ (ii) $(- \frac{4}{5})^{3} = \frac{( - 4 ) ^{3}}{5 ^{3}} = \frac{- 64}{125}$

2.0Some Patterns Related To The Cube

Adding consecutive odd numbers

Observe the following pattern of sum of odd numbers. $1 = 1 = 1^{3}$ $3 + 5 = 8 = 2^{3}$ $7 + 9 + 11 = 27 = 3^{3}$ $13 + 15 + 17 + 19 = 64 = 4^{3}$ $21 + 23 + 25 + 27 + 29 = 125 = 5^{3}$

Only 4 natural numbers, less than 100 are perfect cubes and less than 1000 only 9 numbers are perfect cubes.

Q. How many consecutive odd numbers will be needed to obtain the sum as $10^{3}$ ? Explanation: As we know $4^{3} = 13 + 15 + 17 + 19 = 64$ (4 consecutive odd no. are needed) $5^{3} = 21 + 23 + 25 + 27 + 29 = 125$ (5 consecutive odd no. are needed) . . . $1 0^{3} = 10$ (consecutive odd numbers are needed)

Difference of cubes of consecutive numbers

$2^{3} - 1^{3} = 1 + 2 \times 1 \times 3$ $3^{3} - 2^{3} = 1 + 3 \times 2 \times 3$ $4^{3} - 3^{3} = 1 + 4 \times 3 \times 3$ $⋮$ $(n + 1)^{3} - n^{3} = 1 + (n + 1) \times (n) \times 3$

Q. Using the given pattern, find the value of $5 1^{3} - 5 0^{3}$ . Solution: $∵ (n + 1)^{3} - n^{3} = 1 + (n + 1) (n) (3)$ $∴ 5 1^{3} - 5 0^{3} = 1 + 51 \times 50 \times 3 = 7651$

The square of a negative integer is positive but the cube of a negative integer is negative.

Avoid these mistakes

$(100 + 5)^{3} \neq = 10 0^{3} + 5^{3}$
$(100 - 5)^{3} \neq = 10 0^{3} - 5^{3}$
$2 \times 2 \times 2 = 2^{3}; 2 \times 2 \times 2 \neq = 3^{2}$

3.0Cubes And Their Prime Factors

Consider the prime factorization of the cubes of the numbers. Prime factorization of the cube of numbers $4^{3} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{3} \times 2^{3}$ $1 5^{3} = 3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 = 3^{3} \times 5^{3}$ $1 2^{3} = 1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^{3} \times 2^{3} \times 3^{3}$ It is observed that each prime factor of a number appears three times in the prime factorization of its cube. So, 64, 3375, 1728 are perfect cubes. But in the prime factorization of 500 $500 = 2 \times 2 \times 5 \times 5 \times 5$ There are three 5 's in the product but only two 2 's. So, 500 is not a perfect cube.

Q. Is 11025 a perfect cube? If not, find the smallest natural number by which 11025 must be multiplied so that product is a perfect cube. Explanation: $11025 = 3 \times 3 \times 5 \times 5 \times 7 \times 7$ The prime factors $3, 5, 7$ do not appear in a group of three. Therefore, 11025 is not a perfect cube. To make it a perfect cube, we need one more 3,5 and 7. In that case $11025 \times 3 \times 5 \times 7 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7 = 1157625$ Here, the smallest natural number by which 11025 should be multiplied to make it a perfect cube is 105 .

Volume of a cube $= (side)^{3}$
Side of a cube $= 3 (Volume of a cube)$

Q. Find the volume of a cube with edge 2.2 cm . Solution: Volume $= (edge)^{3} = (2.2)^{3}$ So, $(2.2)^{3} = \frac{2 2 ^{3}}{1 0 ^{3}} = \frac{10648}{1000} = 10.648$ Volume of the cube is $10.648 cm^{3}$ .

4.0Cube Roots

Finding the cube root is the opposite operation of cubing. The symbol for cube root is $3$ i.e., the same as square root but with 3 written in the stroke as shown.

Consider the following:

Statement	Inference	Statement	Inference
$1^{3} = 1$	$31 = 1$	$6^{3} = 216$	$3216 = 3 6^{3} = 6$
$2^{3} = 8$	$38 = 3 2^{3} = 2$	$7^{3} = 343$	$3343 = 3 7^{3} = 7$
$3^{3} = 27$	$327 = 3 3^{3} = 3$	$8^{3} = 512$	$3512 = 3 8^{3} = 8$
$4^{3} = 64$	$364 = 3 4^{3} = 4$	$9^{3} = 729$	$3729 = 3 9^{3} = 9$
$5^{3} = 125$	$3125 = 3 5^{3} = 5$	$1 0^{3} = 1000$	$31000 = 3 1 0^{3} = 10$

Cube roots through prime factorization method Consider 74088. We find its cube root by prime factorization. $74088 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7 = 2^{3} \times 3^{3} \times 7^{3}$ $374088 = 3 2^{3} \times 3^{3} \times 7^{3} = 3 (2 \times 3 \times 7)^{3} = 3 (42)^{3} = 42$

5.0Steps For Finding Cube Roots

Step-1: To find the cube root of a perfect cube, find its prime factors and make them into group of 3.

Step-2 : Pick one factor from each group and multiply them.

Step-3: The product will be the cube root of the given

If a number has one place of decimal, its cube will have 3 places of decimal.

Some common mistakes by students:

$5 + 33 \neq = 315$
$35 - 23 \neq = 32$

Note: These cannot be simplified further.

Q. Find the cube root of 5832 by prime factorization method. Solution: $5832 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$ $5832 = 2^{3} \times 3^{3} \times 3^{3} = (2 \times 3 \times 3)^{3} = (18)^{3}$ $35832 = 3 1 8^{3} = (1 8^{3})^{1/3} = 18$

6.0Cube Roots Of A Cube Number Through Estimation

If you know that the given number is a cube number then following method can be used.

Step-1: Take any cube number say 857375 and start making groups of three digits. Starting from the right most digit of the number.


857	375
$↓$	$↓$
Second group	First group

We get 375 and 857 as two groups of three digits each.

Step-2: First group i.e., 375 will give you the one's digit of unit's digit of the required cube root. The number 375 ends with 5 . We know that 5 comes at the unit's place of a number only when its cube root ends in 5 . So, we get 5 at the unit's place of the cube root.

Step-3: Now take another group, i.e., 857. We know that $9^{3} = 729$ and $1 0^{3} = 1000$ . Also, $729 < 857 < 1000$ . We take the one's place, of the smaller number 729 as the ten's place of the required cube root. So, we get $3857375 = 95$ .

Q. Find the cube root of $17576$ Explanation: The given number is 17576 . Step-1: Form groups of three starting from the right most digit of 17576. $17 \underline{576}$ . In this case one group i.e., 576 has three digits, whereas 17 has only two digits.

Step-2: Take 576. The digit 6 is at its one's place. We take the one's place of the required cube root as 6 .

Step-3: Take the other group, i.e., 17. Cube of 2 is 8 and cube of 3 is 27.17 lies between 8 and 27 . The smaller number among 2 and 3 is 2 . The one's place of 2 is 2 itself. Take 2 as ten's place of the cube root of 17576 . Thus, $317576 = 26$ .

7.0Mind Map

Cubes And Cube Roots

1.0Cubes

If we multiply a number by itself three times, then it is the cube of that number. Cubes in geometry, is a solid figure with 6 faces (which are congruent squares), with 12 edges (which are equal in length) and with 8 vertices. If a cube has an edge of 1 cm , its volume $= 1 \times 1 \times 1 = 1 cm^{3}$ . Similarly, if the edge is 2 cm , its volume $= 2 \times 2 \times 2 = 8 cm^{3}$ . Here 1, 8 are perfect cubes. But 9 is not a perfect cube because there is no natural number which multiplied by itself three times gives 9 . The following table gives the cubes of the first 20 natural numbers.

Number	Cube	Number	Cube
1	1	11	1331
2	8	12	1728
3	27	13	2197
4	64	14	2744
5	125	15	3375
6	216	16	4096
7	343	17	4913
8	512	18	5832
9	729	19	6859
10	1000	20	8000

Some important points

(i) Cube of a rational number $\frac{a}{b}$ is $\frac{a ^{3}}{b ^{3}}$ i.e., $(\frac{a}{b})^{3} = \frac{a ^{3}}{b ^{3}}$ . (ii) Cube of the product of number is equal to the product of their cubes. i.e. $(a \times b)^{3} = a^{3} \times b^{3}$ .

Q. Evaluate: (i) $[4 \times (- 5)]^{3}$ (ii) $(- \frac{4}{5})^{3}$ Explanation: (i) $[4 \times (- 5)]^{3} = 4^{3} \times (- 5)^{3} = 64 \times (- 125) = - 8000$ (ii) $(- \frac{4}{5})^{3} = \frac{( - 4 ) ^{3}}{5 ^{3}} = \frac{- 64}{125}$

2.0Some Patterns Related To The Cube

Adding consecutive odd numbers

Observe the following pattern of sum of odd numbers. $1 = 1 = 1^{3}$ $3 + 5 = 8 = 2^{3}$ $7 + 9 + 11 = 27 = 3^{3}$ $13 + 15 + 17 + 19 = 64 = 4^{3}$ $21 + 23 + 25 + 27 + 29 = 125 = 5^{3}$

Only 4 natural numbers, less than 100 are perfect cubes and less than 1000 only 9 numbers are perfect cubes.

Q. How many consecutive odd numbers will be needed to obtain the sum as $10^{3}$ ? Explanation: As we know $4^{3} = 13 + 15 + 17 + 19 = 64$ (4 consecutive odd no. are needed) $5^{3} = 21 + 23 + 25 + 27 + 29 = 125$ (5 consecutive odd no. are needed) . . . $1 0^{3} = 10$ (consecutive odd numbers are needed)

Difference of cubes of consecutive numbers

$2^{3} - 1^{3} = 1 + 2 \times 1 \times 3$ $3^{3} - 2^{3} = 1 + 3 \times 2 \times 3$ $4^{3} - 3^{3} = 1 + 4 \times 3 \times 3$ $⋮$ $(n + 1)^{3} - n^{3} = 1 + (n + 1) \times (n) \times 3$

Q. Using the given pattern, find the value of $5 1^{3} - 5 0^{3}$ . Solution: $∵ (n + 1)^{3} - n^{3} = 1 + (n + 1) (n) (3)$ $∴ 5 1^{3} - 5 0^{3} = 1 + 51 \times 50 \times 3 = 7651$

The square of a negative integer is positive but the cube of a negative integer is negative.

Avoid these mistakes

$(100 + 5)^{3} \neq = 10 0^{3} + 5^{3}$
$(100 - 5)^{3} \neq = 10 0^{3} - 5^{3}$
$2 \times 2 \times 2 = 2^{3}; 2 \times 2 \times 2 \neq = 3^{2}$

3.0Cubes And Their Prime Factors

Consider the prime factorization of the cubes of the numbers. Prime factorization of the cube of numbers $4^{3} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{3} \times 2^{3}$ $1 5^{3} = 3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 = 3^{3} \times 5^{3}$ $1 2^{3} = 1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^{3} \times 2^{3} \times 3^{3}$ It is observed that each prime factor of a number appears three times in the prime factorization of its cube. So, 64, 3375, 1728 are perfect cubes. But in the prime factorization of 500 $500 = 2 \times 2 \times 5 \times 5 \times 5$ There are three 5 's in the product but only two 2 's. So, 500 is not a perfect cube.

Q. Is 11025 a perfect cube? If not, find the smallest natural number by which 11025 must be multiplied so that product is a perfect cube. Explanation: $11025 = 3 \times 3 \times 5 \times 5 \times 7 \times 7$ The prime factors $3, 5, 7$ do not appear in a group of three. Therefore, 11025 is not a perfect cube. To make it a perfect cube, we need one more 3,5 and 7. In that case $11025 \times 3 \times 5 \times 7 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7 = 1157625$ Here, the smallest natural number by which 11025 should be multiplied to make it a perfect cube is 105 .

Volume of a cube $= (side)^{3}$
Side of a cube $= 3 (Volume of a cube)$

Q. Find the volume of a cube with edge 2.2 cm . Solution: Volume $= (edge)^{3} = (2.2)^{3}$ So, $(2.2)^{3} = \frac{2 2 ^{3}}{1 0 ^{3}} = \frac{10648}{1000} = 10.648$ Volume of the cube is $10.648 cm^{3}$ .

4.0Cube Roots

Finding the cube root is the opposite operation of cubing. The symbol for cube root is $3$ i.e., the same as square root but with 3 written in the stroke as shown.

Consider the following:

Statement	Inference	Statement	Inference
$1^{3} = 1$	$31 = 1$	$6^{3} = 216$	$3216 = 3 6^{3} = 6$
$2^{3} = 8$	$38 = 3 2^{3} = 2$	$7^{3} = 343$	$3343 = 3 7^{3} = 7$
$3^{3} = 27$	$327 = 3 3^{3} = 3$	$8^{3} = 512$	$3512 = 3 8^{3} = 8$
$4^{3} = 64$	$364 = 3 4^{3} = 4$	$9^{3} = 729$	$3729 = 3 9^{3} = 9$
$5^{3} = 125$	$3125 = 3 5^{3} = 5$	$1 0^{3} = 1000$	$31000 = 3 1 0^{3} = 10$

Cube roots through prime factorization method Consider 74088. We find its cube root by prime factorization. $74088 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7 = 2^{3} \times 3^{3} \times 7^{3}$ $374088 = 3 2^{3} \times 3^{3} \times 7^{3} = 3 (2 \times 3 \times 7)^{3} = 3 (42)^{3} = 42$

5.0Steps For Finding Cube Roots

Step-1: To find the cube root of a perfect cube, find its prime factors and make them into group of 3.

Step-2 : Pick one factor from each group and multiply them.

Step-3: The product will be the cube root of the given

If a number has one place of decimal, its cube will have 3 places of decimal.

Some common mistakes by students:

$5 + 33 \neq = 315$
$35 - 23 \neq = 32$

Note: These cannot be simplified further.

Q. Find the cube root of 5832 by prime factorization method. Solution: $5832 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$ $5832 = 2^{3} \times 3^{3} \times 3^{3} = (2 \times 3 \times 3)^{3} = (18)^{3}$ $35832 = 3 1 8^{3} = (1 8^{3})^{1/3} = 18$

6.0Cube Roots Of A Cube Number Through Estimation

If you know that the given number is a cube number then following method can be used.

Step-1: Take any cube number say 857375 and start making groups of three digits. Starting from the right most digit of the number.


857	375
$↓$	$↓$
Second group	First group

We get 375 and 857 as two groups of three digits each.

Step-2: First group i.e., 375 will give you the one's digit of unit's digit of the required cube root. The number 375 ends with 5 . We know that 5 comes at the unit's place of a number only when its cube root ends in 5 . So, we get 5 at the unit's place of the cube root.

Step-3: Now take another group, i.e., 857. We know that $9^{3} = 729$ and $1 0^{3} = 1000$ . Also, $729 < 857 < 1000$ . We take the one's place, of the smaller number 729 as the ten's place of the required cube root. So, we get $3857375 = 95$ .

Q. Find the cube root of $17576$ Explanation: The given number is 17576 . Step-1: Form groups of three starting from the right most digit of 17576. $17 \underline{576}$ . In this case one group i.e., 576 has three digits, whereas 17 has only two digits.

Step-2: Take 576. The digit 6 is at its one's place. We take the one's place of the required cube root as 6 .

Step-3: Take the other group, i.e., 17. Cube of 2 is 8 and cube of 3 is 27.17 lies between 8 and 27 . The smaller number among 2 and 3 is 2 . The one's place of 2 is 2 itself. Take 2 as ten's place of the cube root of 17576 . Thus, $317576 = 26$ .

Related Articles:-

Ratio and Proportion

Algebra

Understanding Elementary Shapes

Decimals

Whole Numbers

Related Articles:-

Ratio and Proportion

Algebra

Understanding Elementary Shapes

Decimals

Whole Numbers

Join ALLEN!

Cubes And Cube Roots

1.0Cubes

2.0Some Patterns Related To The Cube

3.0Cubes And Their Prime Factors

4.0Cube Roots

5.0Steps For Finding Cube Roots

6.0Cube Roots Of A Cube Number Through Estimation

7.0Mind Map

Table of Contents

Join ALLEN!

Cubes And Cube Roots

1.0Cubes

2.0Some Patterns Related To The Cube

3.0Cubes And Their Prime Factors

4.0Cube Roots

5.0Steps For Finding Cube Roots

6.0Cube Roots Of A Cube Number Through Estimation

7.0Mind Map

Table of Contents