Question 13

We can find the cube root of a perfect cube by looking at its digits and grouping them.

Step 1 — Understanding unit digits We first look at the last digit of the number. This tells us the last digit of its cube root. Let us see the pattern for unit digits of cubes: $1^3 = 1$ $2^3 = 8$ $3^3 = 27$ $4^3 = 64$ $5^3 = 125$ $6^3 = 216$ $7^3 = 343$ $8^3 = 512$ $9^3 = 729$ $10^3 = 1000$ If a number ends in 1, its cube root ends in 1. If a number ends in 8, its cube root ends in 2. If a number ends in 7, its cube root ends in 3. If a number ends in 4, its cube root ends in 4. If a number ends in 5, its cube root ends in 5. If a number ends in 6, its cube root ends in 6. If a number ends in 3, its cube root ends in 7. If a number ends in 2, its cube root ends in 8. If a number ends in 9, its cube root ends in 9. If a number ends in 0, its cube root ends in 0.

Step 2 — Finding the ten's digit We group the digits of the number from the right. We make groups of three digits. The first group from the left determines the ten's digit. We find the largest single digit number whose cube is less than or equal to this first group. This number will be the ten's digit of our cube root.

Step 3 — Cube root of 1331 Let us find the cube root of 1331. The number ends in 1. So, the unit's digit of its cube root is 1. We group the digits from the right. The groups are 1 and 331. The first group from the left is 1. We find the largest number whose cube is less than or equal to 1. $1^3 = 1$ So, the ten's digit of the cube root is 1. Combining the digits, the cube root is 11.

$\text{Unit's digit of } \sqrt[3]{1331} = 1 \quad (\text{since } 1331 \text{ ends in } 1)$

$\text{First group from left is } 1$

$1^3 = 1$

$\text{Ten's digit of } \sqrt[3]{1331} = 1$

$\boxed{\sqrt[3]{1331} = 11}$

Step 4 — Cube root of 4913 Let us find the cube root of 4913. The number ends in 3. So, the unit's digit of its cube root is 7 (since $7^3 = 343$). We group the digits from the right. The groups are 4 and 913. The first group from the left is 4. We find the largest number whose cube is less than or equal to 4. $1^3 = 1 \quad \text{and} \quad 2^3 = 8$ Since $1^3 \le 4 < 2^3$, the ten's digit is 1. Combining the digits, the cube root is 17.

$\text{Unit's digit of } \sqrt[3]{4913} = 7 \quad (\text{since } 4913 \text{ ends in } 3)$

$\text{First group from left is } 4$

$1^3 = 1 \quad \text{and} \quad 2^3 = 8$

$\text{Since } 1^3 \le 4 < 2^3, \text{ ten's digit is } 1$

$\boxed{\sqrt[3]{4913} = 17}$

Step 5 — Cube root of 12167 Let us find the cube root of 12167. The number ends in 7. So, the unit's digit of its cube root is 3 (since $3^3 = 27$). We group the digits from the right. The groups are 12 and 167. The first group from the left is 12. We find the largest number whose cube is less than or equal to 12. $2^3 = 8 \quad \text{and} \quad 3^3 = 27$ Since $2^3 \le 12 < 3^3$, the ten's digit is 2. Combining the digits, the cube root is 23.

$\text{Unit's digit of } \sqrt[3]{12167} = 3 \quad (\text{since } 12167 \text{ ends in } 7)$

$\text{First group from left is } 12$

$2^3 = 8 \quad \text{and} \quad 3^3 = 27$

$\text{Since } 2^3 \le 12 < 3^3, \text{ ten's digit is } 2$

$\boxed{\sqrt[3]{12167} = 23}$

Step 6 — Cube root of 32768 Let us find the cube root of 32768. The number ends in 8. So, the unit's digit of its cube root is 2 (since $2^3 = 8$). We group the digits from the right. The groups are 32 and 768. The first group from the left is 32. We find the largest number whose cube is less than or equal to 32. $3^3 = 27 \quad \text{and} \quad 4^3 = 64$ Since $3^3 \le 32 < 4^3$, the ten's digit is 3. Combining the digits, the cube root is 32.

$\text{Unit's digit of } \sqrt[3]{32768} = 2 \quad (\text{since } 32768 \text{ ends in } 8)$

$\text{First group from left is } 32$

$3^3 = 27 \quad \text{and} \quad 4^3 = 64$

$\text{Since } 3^3 \le 32 < 4^3, \text{ ten's digit is } 3$

$\boxed{\sqrt[3]{32768} = 32}$

Answer

(i) The cube root of 1331 is 11. (ii) The cube root of 4913 is 17. (iii) The cube root of 12167 is 23. (iv) The cube root of 32768 is 32.