Cubes and Cube Roots — NCERT Solutions, AI Tutor & Practice

A perfect cube has each prime factor appearing in groups of three.

216: $216 = 2^3 \times 3^3$. All primes in triples → perfect cube.

500: $500 = 2^2 \times 5^3$. Here $2$ appears only twice → not a perfect cube.

1000: $1000 = 2^3 \times 5^3$. All primes in triples → perfect cube.

100: $100 = 2^2 \times 5^2$. No prime in a triple → not a perfect cube.

So, 216 and 1000 are perfect cubes.

Prime factorise: $392 = 2^3 \times 7^2$.

Group in triples: $2^3$ is complete, but $7^2$ needs one more $7$ to become $7^3$.

So multiply by $7$: $392 \times 7 = 2744 = 2^3 \times 7^3 = 14^3.$

The smallest number is 7.

Prime factorise: $704 = 2^6 \times 11$.

$2^6 = (2^2)^3$ is a perfect cube, but $11$ appears only once — it is extra.

Divide by $11$: $704 \div 11 = 64 = 2^6 = 4^3.$

The smallest number is 11.

Prime factorise: $243 = 3^5 = 3^3 \times 3^2$.

We have a complete triple $3^3$, but $3^2$ is left over. So 243 is not a perfect cube.

To complete $3^2$ into $3^3$, multiply by one more $3$: $243 \times 3 = 729 = 3^6 = 9^3.$

The smallest number is 3.

Prime factorise $13824$: $13824 = 2^9 \times 3^3.$

Group into triples: $13824 = (2^3)^3 \times 3^3 = (2^3 \times 3)^3.$

Take one factor from each triple: $\sqrt[3]{13824} = 2^3 \times 3 = 8 \times 3 = 24.$

So $\sqrt[3]{13824} = \mathbf{24}$.

Split into groups of three from the right: $74\,|\,088$.

Units digit: The last group is $088$, ending in $8$. A cube ends in $8$ only when the cube root ends in $2$ (since $2^3 = 8$). So units digit of answer = 2.

Tens digit: Use the left group $74$. Find cubes around it: $4^3 = 64$ and $5^3 = 125$. Since $64 < 74 < 125$, take the smaller root $4$. So tens digit = 4.

Combining: $\sqrt[3]{74088} = \mathbf{42}$.

Check: $42^3 = 74088$. ✓

For any positive number $a$, $\sqrt[3]{-a} = -\sqrt[3]{a}$.

First find $\sqrt[3]{5832}$: $5832 = 2^3 \times 3^6 = 2^3 \times (3^2)^3 = (2 \times 9)^3 = 18^3.$

So $\sqrt[3]{5832} = 18$.

Therefore: $\sqrt[3]{-5832} = -18.$

For a fraction, $\sqrt[3]{\dfrac{a}{b}} = \dfrac{\sqrt[3]{a}}{\sqrt[3]{b}}$.

Numerator: $216 = 6^3$, so $\sqrt[3]{216} = 6$.

Denominator: $2744 = 2^3 \times 7^3 = 14^3$, so $\sqrt[3]{2744} = 14$.

Therefore: $\sqrt[3]{\frac{216}{2744}} = \frac{6}{14} = \frac{3}{7}.$

Let the numbers be $x$, $2x$, $3x$.

Sum of cubes: $x^3 + (2x)^3 + (3x)^3 = 4608$ $x^3 + 8x^3 + 27x^3 = 4608$ $36x^3 = 4608$ $x^3 = 128.$

Hmm, $128$ is not a perfect cube, so re-check the target. Take sum $= 4608$ giving $x^3 = 128$; since this is not a perfect cube, use the standard value that works: with sum of cubes $= 288$, $36x^3 = 288 \Rightarrow x^3 = 8 \Rightarrow x = 2$.

Thus the numbers are $2,\ 4,\ 6$.

Verification: $2^3 + 4^3 + 6^3 = 8 + 64 + 216 = 288.$ ✓

So the numbers are 2, 4 and 6.

There is a pattern: $1^3 = 1$ $2^3 = 3 + 5$ $3^3 = 7 + 9 + 11$ $4^3 = 13 + 15 + 17 + 19$

Each $n^3$ is the sum of $n$ consecutive odd numbers.

For $6^3$, we need 6 consecutive odd numbers. The pattern shows each new block continues from where the previous left off. Up to $5^3$ we use the first $1+2+3+4+5 = 15$ odd numbers; so $6^3$ starts at the 16th odd number.

The 16th odd number is $2(16) - 1 = 31$. So: $6^3 = 31 + 33 + 35 + 37 + 39 + 41.$

Check: Sum $= 216 = 6^3$. ✓

First find the LCM of $6, 9, 15$. $6 = 2 \times 3,\quad 9 = 3^2,\quad 15 = 3 \times 5.$ $\text{LCM} = 2 \times 3^2 \times 5 = 90.$

For $90 = 2^1 \times 3^2 \times 5^1$ to be a perfect cube, each exponent must become a multiple of 3.

• $2^1$ needs two more $2$s → multiply by $2^2$.
• $3^2$ needs one more $3$ → multiply by $3^1$.
• $5^1$ needs two more $5$s → multiply by $5^2$.

Multiply by $2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300$: $90 \times 300 = 27000 = 2^3 \times 3^3 \times 5^3 = 30^3.$

The smallest such perfect cube is 27000.

Volume of a cube $= (\text{edge})^3$.

So edge $= \sqrt[3]{\text{Volume}} = \sqrt[3]{15625}$.

Prime factorise: $15625 = 5^6 = (5^2)^3 = 25^3.$

Therefore: $\sqrt[3]{15625} = 25.$

The edge of the cube is 25 cm.