Exercise 6.5 — Cube Root Methods

Cube roots by prime factorisation and estimating cube roots.

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Exercise 6.5 – Cube Roots: Prime Factorisation & Estimation Methods

Exercise 6.5 from Chapter 6, "Square Roots and Cube Roots," of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) introduces cube roots and teaches two practical methods to find them — the prime factorisation method for exact answers, and the estimation method for quickly finding the cube root of larger perfect cubes without full factorisation. The exercise also includes a true/false conceptual quiz and a classic problem about numbers that are both perfect squares and perfect cubes.

This exercise is the natural follow-up to Exercise 6.4 on cubes and perfect cubes, where you learned how to test whether a number is a perfect cube using prime factorisation. Here, that same factorisation is used to actually extract the cube root.

What Is a Cube Root?

If a perfect cube can be written as the product of three equal integers, that integer is called the cube root of the perfect cube. In other words, if y = x³, then x is the cube root of y, written using the radical symbol as shown below.

If y = x³, then ∛y = x

For example, since 2³ = 8, the cube root of 8 is 2 (written ∛8 = 2). Similarly, since 5³ = 125, the cube root of 125 is 5 (written ∛125 = 5). The cube root "undoes" the cubing operation.

Cubic Numbers and Their Cube Roots (1 to 20)

This reference table is extremely useful for both the prime factorisation and estimation methods, since recognising these values instantly speeds up calculations:

Cubic NumberCube RootCubic NumberCube Root
11133111
82172812
273219713
644274414
1255337515
2166409616
3437491317
5128583218
7299685919
100010800020

Units Digit Pattern — Cube to Cube Root

Just like cubes have a predictable units-digit pattern, cube roots follow the reverse of that same pattern. This table is the key to the estimation method used later in this exercise:

Units Digit in the Cubic NumberUnits Digit in the Cube Root
11
28
37
44
55
66
73
82
99
00
💡 Key Insight: Notice that 2 and 8 swap, and 3 and 7 swap — this is exactly the reverse of the cube units-digit table from Exercise 6.4. All other digits (0, 1, 4, 5, 6, 9) map to themselves in both directions.

Exercise 6.5 – Solved Questions, Step by Step

Question 1
Find the Cube Root by Prime Factorisation Method
Prime Factorisation Method

In this method, write the prime factorisation of the number, group the identical prime factors into sets of three, and take one factor from each group to form the cube root.

(i) Cube Root of 343

7343
749
 7

343 = 7 × 7 × 7 — one complete group of three 7s. Taking one factor from this group gives the cube root.

∛343 = 7

(ii) Cube Root of 729

3729
3243
381
327
39
 3

729 = 3 × 3 × 3 × 3 × 3 × 3 — six 3s, forming two complete groups of three. Taking one factor (3) from each group and multiplying gives the cube root.

∛729 = 3 × 3 = 9

(iii) Cube Root of 1331

111331
11121
 11

1331 = 11 × 11 × 11 — one complete group of three 11s.

∛1331 = 11

(iv) Cube Root of 2744

22744
21372
2686
7343
749
 7

2744 = 2 × 2 × 2 × 7 × 7 × 7 — two complete groups of three (one group of 2s and one group of 7s). Taking one factor from each group and multiplying gives the cube root.

∛2744 = 2 × 7 = 14
Question 2
Find the Cube Root Through Estimation
Estimation Method

The estimation method is a fast way to find the cube root of perfect cubes (usually up to 4-5 digits) without full prime factorisation. The procedure is:

  1. Step 1: Starting from the right, split the number into groups of three digits each. The rightmost group is the "first group," and whatever remains is the "second group."
  2. Step 2: Use the units digit of the first group to find the units digit of the cube root, using the cube-root units-digit table.
  3. Step 3: Look at the second group. Find the two consecutive cubic numbers (from the 1-20 table) between which this second-group number lies. The cube root of the smaller of those two cubic numbers gives the tens digit of the answer.

(i) Cube Root of 512

512 has only 3 digits, so it forms a single group: 512. The units digit of this group is 2, and according to the units-digit table, a cube ending in 2 has a cube root ending in 8.

∛512 = 8

(ii) Cube Root of 2197

Splitting into groups of three from the right: first group = 197, second group = 2.

  • The units digit of the first group (197) is 7. From the table, a cube ending in 7 has a cube root ending in 3 — so the units digit of the answer is 3.
  • The second group is 2. Since 1³ = 1 and 2³ = 8, the number 2 lies between 1 and 8 (i.e., 1 < 2 < 8). The smaller cube root here is 1, so the tens digit of the answer is 1.
∛2197 = 13

(iii) Cube Root of 3375

Splitting into groups: first group = 375, second group = 3.

  • The units digit of 375 is 5. A cube ending in 5 has a cube root ending in 5 — so the units digit of the answer is 5.
  • The second group is 3. Since 1³ = 1 and 2³ = 8, the number 3 lies between 1 and 8 (1 < 3 < 8). The smaller cube root is 1, so the tens digit of the answer is 1.
∛3375 = 15

(iv) Cube Root of 5832

Splitting into groups: first group = 832, second group = 5.

  • The units digit of 832 is 2. A cube ending in 2 has a cube root ending in 8 — so the units digit of the answer is 8.
  • The second group is 5. Since 1³ = 1 and 2³ = 8, the number 5 lies between 1 and 8 (1 < 5 < 8). The smaller cube root is 1, so the tens digit of the answer is 1.
∛5832 = 18
📐 Why Estimation Works: The estimation method works only for perfect cubes with up to 6 digits, because the "second group" (the leftover digits after removing the rightmost 3) will always be a number from 1 to 999, which corresponds to a cube root from 1 to 9 for the tens digit.
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Question 3
True or False — Conceptual Statements About Cubes
Conceptual Understanding

This question tests deeper understanding of how cubes behave, often catching common misconceptions. Each statement is evaluated below:

#StatementAnswer
(i)Cube of an even number is an odd numberFalse
(ii)A perfect cube may end with two zerosFalse
(iii)If a number ends with 5, then its cube ends with 5True
(iv)Cube of a number ending with zero has three zeros at its rightTrue
(v)The cube of a single digit number is a single digit numberFalse
(vi)There is no perfect cube which ends with 8False
(vii)The cube of a two digit number may be a three digit numberFalse

Why Each Answer Is Correct

  • (i) False: The cube of an even number is always even (e.g., 2³ = 8), not odd.
  • (ii) False: A perfect cube can only end in zeros in multiples of three (0, 3, 6, 9...). Ending in exactly two zeros is impossible — for example, a number ending in one zero must have its cube end in three zeros.
  • (iii) True: From the units-digit table, a number ending in 5 always has a cube ending in 5 (e.g., 5³ = 125, 15³ = 3375).
  • (iv) True: If a number ends with one zero (like 10, 20, 30), its cube ends with three zeros (10³ = 1000, 20³ = 8000).
  • (v) False: While 1³ = 1 and 2³ = 8 are single digits, from 3³ = 27 onwards, cubes of single-digit numbers become multi-digit numbers.
  • (vi) False: 12³ = 1728 ends in 8, so perfect cubes ending in 8 definitely exist (any number ending in 2 has a cube ending in 8).
  • (vii) False: The smallest two-digit number is 10, and 10³ = 1000, which already has 4 digits. So the cube of any two-digit number has at least 4 digits, never just 3.
Question 4
Find a Two-Digit Number That Is Both a Perfect Square and a Perfect Cube
LCM of Exponents

This is a classic problem that combines the ideas of squares and cubes. A number is a perfect square if it can be written as x², and a perfect cube if it can be written as x³. For a number to be both, it must be expressible as x raised to a power that is a common multiple of 2 and 3.

  1. Step 1: The LCM of the exponents 2 and 3 is 6. So any number of the form x⁶ will be both a perfect square (since x⁶ = (x³)²) and a perfect cube (since x⁶ = (x²)³).
  2. Step 2: Try x = 1: x⁶ = 1 × 1 × 1 × 1 × 1 × 1 = 1 (a one-digit number).
  3. Step 3: Try x = 2: x⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64 (a two-digit number).
  4. Step 4: Try x = 3: x⁶ = 3 × 3 × 3 × 3 × 3 × 3 = 729 (a three-digit number — already too large).
The only two-digit number that is both a perfect square and a perfect cube = 64 (= 8² = 4³)
📐 Verify It Yourself: 64 = 8 × 8 (perfect square) and 64 = 4 × 4 × 4 (perfect cube). Both checks confirm 64 is the answer — and since x = 1 gives a one-digit number and x = 3 already jumps to a three-digit number, 64 is the only two-digit possibility.

Common Mistakes to Avoid in Exercise 6.5

  • Forgetting to take only one factor per group: In the prime factorisation method, each group of three identical factors contributes only one factor to the cube root — don't carry over all three.
  • Misgrouping digits in the estimation method: Always start grouping digits in threes from the right-hand side (units place), not from the left.
  • Mixing up the cube units-digit table and the cube-root units-digit table: They look similar but are reverses of each other — always check which direction you need (cube → root, or root → cube).
  • Assuming the estimation method works for non-perfect-cubes: This method only gives a correct answer when the original number is already a perfect cube. For non-perfect cubes, prime factorisation or a calculator is needed.
  • Incorrectly evaluating true/false statements by testing only one example: A statement like "the cube of a single-digit number is a single-digit number" might seem true for 1 and 2, but a single counterexample (like 3³ = 27) makes the whole statement false.

What Exercise 6.5 Prepares You For

Cube roots, along with the square roots covered earlier in this chapter, form the foundation for simplifying expressions involving exponents and powers in later exercises. The LCM-of- exponents idea from Question 4 directly connects to Exponents and Powers, where laws of exponents are studied in more depth.

These skills also carry forward into Class 9 and Class 10, especially when working with real numbers and irrational numbers, and when simplifying algebraic expressions involving cube identities such as a³ + b³ and a³ − b³ in Polynomials. Students preparing for CBSE, Telangana, or Andhra Pradesh board exams should be confident with both the prime factorisation and estimation methods, since either can appear depending on the size of the number given.

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