Cube and Cube Roots

Q1: Multiple Choice Type:

i. The cube root of 0.000027 is:

Step 1: Write 0.000027 in fractional form: \[ 0.000027 = \frac{27}{1000000} \]Step 2: Find cube root of numerator and denominator separately: \[ \sqrt[3]{27} = 3,\quad \sqrt[3]{1000000} = 100 \]So, \[ \sqrt[3]{0.000027} = \frac{3}{100} = 0.03 \]Answer: a. 0.03

ii. The cube root of -0.064 is:

Step 1: Cube root of a negative number is negative
Step 2: Write -0.064 as a fraction: \[ -0.064 = -\frac{64}{1000} \\ \sqrt[3]{64} = 4,\quad \sqrt[3]{1000} = 10 \]So, \[ \sqrt[3]{-0.064} = -\frac{4}{10} = -0.4 \]Answer: d. -0.4

Q2: Find the cube-roots of:

i. 64

Step 1: Prime factorise 64: \[ 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 \] Step 2: Group in triples: \[ 2^6 = (2^3)^2 \\ \Rightarrow \sqrt[3]{64} = 2^2 = 4 \] Answer: 4

ii. 343

Step 1: Prime factorise: \[ 343 = 7 \times 7 \times 7 = 7^3 \\ \Rightarrow \sqrt[3]{343} = 7 \] Answer: 7

iii. 729

Step 1: Prime factorise: \[ 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 \\ \Rightarrow \sqrt[3]{729} = 3^2 = 9 \] Answer: 9

iv. 1728

Step 1: Prime factorise: \[ 1728 = 2^6 \times 3^3 \\ \Rightarrow \sqrt[3]{1728} = 2^2 \times 3 = 4 \times 3 = 12 \] Answer: 12

v. 9261

Step 1: Prime factorise: \[ 9261 = 3 \times 3 \times 3 \times 7 \times 7 \times 7 = 3^3 \times 7^3 \\ \Rightarrow \sqrt[3]{9261} = 3 \times 7 = 21 \] Answer: 21

vi. 4096

Step 1: Prime factorise: \[ 4096 = 2^{12} \\ \Rightarrow \sqrt[3]{4096} = 2^4 = 16 \quad \text{(since } 2^3 \times 2^3 \times 2^3 \times 2^3 = 4096) \] Answer: 16

vii. 8000

Step 1: Prime factorise: \[ 8000 = 2^6 \times 5^3 \\ \Rightarrow \sqrt[3]{8000} = 2^2 \times 5 = 4 \times 5 = 20 \] Answer: 20

viii. 3375

Step 1: Prime factorise: \[ 3375 = 3^3 \times 5^3 \\ \Rightarrow \sqrt[3]{3375} = 3 \times 5 = 15 \] Answer: 15

Q3: Find the cube-roots of:

i. \(\frac{27}{64}\)

Step 1: Prime factorise numerator and denominator: \[ 27 = 3^3,\quad 64 = 2^6 = (2^3)^2 \\ \sqrt[3]{\frac{27}{64}} = \frac{\sqrt[3]{3^3}}{\sqrt[3]{2^6}} = \frac{3}{2^2} = \frac{3}{4} \]Answer: \(\frac{3}{4}\)

ii. \(\frac{125}{216}\)

Step 1: Prime factorise: \[ 125 = 5^3,\quad 216 = 2^3 \times 3^3 \\ \sqrt[3]{\frac{125}{216}} = \frac{5}{2 \times 3} = \frac{5}{6} \]Answer: \(\frac{5}{6}\)

iii. \(\frac{343}{512}\)

Step 1: Prime factorise: \[ 343 = 7^3,\quad 512 = 2^9 = (2^3)^3 \\ \sqrt[3]{\frac{343}{512}} = \frac{7}{2^3} = \frac{7}{8} \]Answer: \(\frac{7}{8}\)

iv. \(64 \times 729\)

Step 1: Prime factorise: \[ 64 = 2^6,\quad 729 = 3^6 \\ \Rightarrow 64 \times 729 = 2^6 \times 3^6 = (2^2 \times 3^2)^3 = 36^3 \\ \sqrt[3]{64 \times 729} = 36 \]Answer: 36

v. \(64 \times 27\)

Step 1: Prime factorise: \[ 64 = 2^6,\quad 27 = 3^3 \\ \Rightarrow 64 \times 27 = 2^6 \times 3^3 = (2^2)^3 \times 3^3 = (4 \times 3)^3 = 12^3 \\ \sqrt[3]{64 \times 27} = 12 \]Answer: 12

vi. \(729 \times 8000\)

Step 1: Prime factorise: \[ 729 = 3^6,\quad 8000 = 2^6 \times 5^3 \\ \Rightarrow 729 \times 8000 = 2^6 \times 3^6 \times 5^3 = (2^2 \times 3^2 \times 5)^3 = 180^3 \\ \sqrt[3]{729 \times 8000} = 180 \]Answer: 180

vii. \(3375 \times 512\)

Step 1: Prime factorise: \[ 3375 = 3^3 \times 5^3,\quad 512 = 2^9 \\ \Rightarrow 3375 \times 512 = 2^9 \times 3^3 \times 5^3 = (2^3 \times 3 \times 5)^3 = (8 \times 3 \times 5)^3 = 120^3 \\ \sqrt[3]{3375 \times 512} = 120 \]Answer: 120

Q4: Find the cube-roots of:

i. \(-216\)

Step 1: \(216 = 2^3 \times 3^3\) \[ \sqrt[3]{-216} = -\sqrt[3]{216} = – (2 \times 3) = -6 \] Answer: -6

ii. \(-512\)

Step 1: \(512 = 2^9 = (2^3)^3\) \[ \sqrt[3]{-512} = -2^3 = -8 \] Answer: -8

iii. \(-1331\)

Step 1: \(1331 = 11^3\) \[ \sqrt[3]{-1331} = -11 \] Answer: -11

iv. \(-\frac{27}{125}\)

Step 1: \(27 = 3^3,\quad 125 = 5^3\) \[ \sqrt[3]{-\frac{27}{125}} = -\frac{3}{5} \] Answer: \(-\frac{3}{5}\)

v. \(\frac{-64}{343}\)

Step 1: \(64 = 2^6,\quad 343 = 7^3\) \[ \sqrt[3]{\frac{-64}{343}} = -\frac{2^2}{7} = -\frac{4}{7} \] Answer: \(-\frac{4}{7}\)

vi. \(-\frac{512}{343}\)

Step 1: \(512 = 2^9,\quad 343 = 7^3\) \[ \sqrt[3]{-\frac{512}{343}} = -\frac{2^3}{7} = -\frac{8}{7} \] Answer: \(-\frac{8}{7}\)

vii. \(-2197\)

Step 1: \(2197 = 13^3\) \[ \sqrt[3]{-2197} = -13 \] Answer: -13

viii. \(-5832\)

Step 1: \(5832 = 2^3 \times 3^6\) \[ \sqrt[3]{-5832} = – (2 \times 3^2) = – (2 \times 9) = -18 \] Answer: -18

ix. \(-2744000\)

Step 1: Ignore the negative sign for now and factorise 2744000
We write: \[ 2744000 = 2744 \times 1000 \]Now factorise: \[ 2744 = 2^3 \times 7^3 \quad (\text{since } 2744 = 14^3 = 2^3 \times 7^3) \\ 1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3 \]So, \[ 2744000 = 2^3 \times 7^3 \times 2^3 \times 5^3 = 2^6 \times 5^3 \times 7^3 \]Group them: \[ = (2^2 \times 5 \times 7)^3 = (4 \times 5 \times 7)^3 = (140)^3 \]Now include the negative sign: \[ \sqrt[3]{-2744000} = -140 \]Answer: -140

Q5: Find the cube-roots of:

i. 2.744

Step 1: Convert to fraction: \[ 2.744 = \frac{2744}{1000} \] Step 2: Prime factorise: \[ 2744 = 14^3 = 2^3 \times 7^3,\quad 1000 = 10^3 = 2^3 \times 5^3 \\ \sqrt[3]{\frac{2744}{1000}} = \frac{14}{10} = 1.4 \] Answer: 1.4

ii. 9.261

Step 1: Convert to fraction: \[ 9.261 = \frac{9261}{1000} \\ 9261 = (3 \times 7)^3 = 21^3,\quad 1000 = 10^3 \\ \sqrt[3]{9.261} = \sqrt[3]{\frac{9261}{1000}} = \frac{21}{10} = 2.1 \] Answer: 2.1

iii. 0.000027

Step 1: Convert to fraction: \[ 0.000027 = \frac{27}{1000000} \\ 27 = 3^3,\quad 1000000 = (10^6) = (10^2)^3 = 100^3 \\ \sqrt[3]{0.000027} = \frac{3}{100} = 0.03 \] Answer: 0.03

iv. -0.512

Step 1: Use cube root of positive 0.512 first: \[ 0.512 = \frac{512}{1000},\quad 512 = 8^3,\quad 1000 = 10^3 \\ \sqrt[3]{-0.512} = -\frac{8}{10} = -0.8 \] Answer: -0.8

v. -15.625

Step 1: Convert to fraction: \[ 15.625 = \frac{15625}{1000} \\ 15625 = 25^3 = (5^2)^3 = 5^6,\quad 1000 = 10^3 = 2^3 \times 5^3 \\ \sqrt[3]{-15.625} = -\sqrt[3]{\frac{15625}{1000}} = -\frac{25}{10} = -2.5 \] Answer: -2.5

vi. \(-125 \times 1000\)

Step 1: Multiply: \[ -125 \times 1000 = -125000 \\ 125 = 5^3,\quad 1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3 \\ \Rightarrow 125000 = 2^3 \times 5^6 \\ \sqrt[3]{-125000} = – (2 \times 5^2) = – (2 \times 25) = -50 \] Answer: -50

Q6: Find the smallest number by which 26244 should be divided so that the quotient is a perfect cube.

Step 1: Perform prime factorisation of 26244 \[ 26244 \div 2 = 13122 \\ 13122 \div 2 = 6561 \\ 6561 \div 3 = 2187 \\ 2187 \div 3 = 729 \\ 729 \div 3 = 243 \\ 243 \div 3 = 81 \\ 81 \div 3 = 27 \\ 27 \div 3 = 9 \\ 9 \div 3 = 3 \\ 3 \div 3 = 1 \]So, \[ 26244 = 2^2 \times 3^8 \]Step 2: To form a perfect cube, the power of each prime factor must be a multiple of 3.
→ \(2^2\) is not a cube (we need \(2^3\))
→ \(3^8\) is not a cube (we need \(3^6\) or \(3^9\))
To adjust:
→ Remove one \(3^2\) (since \(8 – 6 = 2\)), and one \(2^2\) to get cube form: \[ \text{Divide by } 2^2 \times 3^2 = 4 \times 9 = 36 \]So, \[ \frac{26244}{36} = 729 \\ 729 = 3^6 = (3^2)^3 = 9^3 \\ \Rightarrow \text{Perfect cube} \]Answer: 36

Q7: What is the least number by which 30375 should be multiplied to get a perfect cube?

Step 1: Perform prime factorisation of 30375 \[ 30375 \div 5 = 6075 \\ 6075 \div 5 = 1215 \\ 1215 \div 5 = 243 \\ 243 \div 3 = 81 \\ 81 \div 3 = 27 \\ 27 \div 3 = 9 \\ 9 \div 3 = 3 \\ 3 \div 3 = 1 \]So, \[ 30375 = 3^5 \times 5^3 \]Step 2: To form a perfect cube, all powers of prime factors must be multiples of 3.
→ \(3^5\) is not a perfect cube (we need \(3^6\), so multiply by one more 3)
→ \(5^3\) is already a perfect cube
Step 3: Multiply by 3 to make all powers multiples of 3: \[ 30375 \times 3 = 91125 \\ 91125 = 3^6 \times 5^3 = (3^2)^3 \times 5^3 = (9 \times 5)^3 = 45^3 \\ \Rightarrow \text{Perfect cube} \]Answer: 3

Q8: Find the cube-roots of:

i. \(700 \times 2 \times 49 \times 5\)

Step 1: Multiply the numbers:
\(700 = 2^2 \times 5^2 \times 7\),
\(49 = 7^2\),
Total product = \(700 \times 2 \times 49 \times 5 = (2^3 \times 5^3 \times 7^3)\) \[ = (2 \times 5 \times 7)^3 = (70)^3 \\ \Rightarrow \sqrt[3]{700 \times 2 \times 49 \times 5} = 70 \] Answer: 70

ii. \(-216 \times 1728\)

Step 1: Prime factorise:
\(216 = 2^3 \times 3^3,\quad 1728 = 2^6 \times 3^3\) \[ -216 \times 1728 = – (2^9 \times 3^6) = – (2^3 \times 3^2)^3 = – (8 \times 9)^3 = -72 \\ \sqrt[3]{-216 \times 1728} = -72 \] Answer: -72

iii. \(-64 \times -125\)

Step 1: \(-64 \times -125 = 8000\) \[ 8000 = 2^6 \times 5^3 = (2^2 \times 5)^3 = (4 \times 5)^3 = 20^3 \\ \sqrt[3]{8000} = 20 \] Answer: 20

iv. \(-\frac{27}{343}\)

Step 1: Prime factors: \[ 27 = 3^3,\quad 343 = 7^3 \Rightarrow \sqrt[3]{-\frac{27}{343}} = -\frac{3}{7} \] Answer: \(-\frac{3}{7}\)

v. \(\frac{729}{-1331}\)

Step 1: Prime factors: \[ 729 = 3^6 = (3^2)^3,\quad 1331 = 11^3 \Rightarrow \sqrt[3]{\frac{729}{-1331}} = -\frac{9}{11} \] Answer: \(-\frac{9}{11}\)

vi. 250.047

Step 1: Check if it’s a cube of a decimal:
Try \(6.3^3 = 250.047\) \[ 6.3 \times 6.3 = 39.69 \quad \text{then} \quad 39.69 \times 6.3 = 250.047 \\ \Rightarrow \sqrt[3]{250.047} = 6.3 \] Answer: 6.3

vii. \(-175616\)

Step 1: Prime factorise 175616: \[ 175616 = 2^5 \times 11^3 \Rightarrow (2 \times 11)^3 \times 2^2 = 22^3 \times 4 \text{ → not a perfect cube} \] Actually, 175616 = \(56^3\)
So, \[ \sqrt[3]{-175616} = -56 \] Answer: -56