NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 1 — Figure it Out (Page 16)#

Cube root of 27000. We write 27000 as $27 \times 1000$ and factorise:

$27000 = 27 \times 1000$
$27000 = (3 \times 3 \times 3) \times (10 \times 10 \times 10)$
$27000 = (3 \times 10) \times (3 \times 10) \times (3 \times 10)$
$27000 = 30 \times 30 \times 30$

$\sqrt[3]{27000} = 30$

Cube root of 10648. We divide by 2 until we can’t:

$10648 \div 2 = 5324$
$5324 \div 2 = 2662$
$2662 \div 2 = 1331$

and $1331 = 11 \times 11 \times 11$. So:

$10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11$
$10648 = (2 \times 11) \times (2 \times 11) \times (2 \times 11)$
$10648 = 22 \times 22 \times 22$

$\sqrt[3]{10648} = 22$

∴ $\sqrt[3]{27000} = $ 30 and $\sqrt[3]{10648} = $ 22.

We factorise 1323 into primes:

$1323 = 3 \times 3 \times 3 \times 7 \times 7$

The 3s already form a group of three, but there are only two 7s. We need one more 7 to complete a group of three:

$1323 \times 7 = 3 \times 3 \times 3 \times 7 \times 7 \times 7$
$1323 \times 7 = (3 \times 7) \times (3 \times 7) \times (3 \times 7)$
$1323 \times 7 = 21 \times 21 \times 21$
$1323 \times 7 = 9261$

∴ We multiply 1323 by 7 to get the cube number $9261 = 21^3$.

False. An odd number multiplied by an odd number is odd, so odd × odd × odd is odd. For example, $3^3 = 27$ and $5^3 = 125$ are both odd.

False. $2^3 = 8$ ends with 8. In fact, the cube of every number ending in 2 ends with 8, for example $12^3 = 1728$.

False. The smallest 2-digit number is 10, and its cube already has 4 digits:

$10^3 = 1000$

So the cube of any 2-digit number has at least 4 digits.

False. The largest 2-digit number is 99, and its cube has only 6 digits:

$99^3 = 970299$

So the cube of any 2-digit number has at most 6 digits.

False. The cube $8 = 2^3$ has the factors 1, 2, 4 and 8, which is 4 factors, an even number. Only perfect squares have an odd number of factors, so a cube has an odd number of factors only when it is also a square, like $64 = 4^3 = 8^2$.

We use two clues for each number:

• The last digit. Cubes of 0 to 9 end in different digits, so the last digit of the cube gives the last digit of the cube root:

• The size. Leaving out the last three digits, we see between which cubes of 1, 2, 3, … the rest lies. That gives the tens digit.

1331. It ends in 1, so the cube root ends in 1. Leaving out 331 leaves 1, and $1^3 = 1$, so the tens digit is 1. The cube root is 11. Check: $11^3 = 1331$.

4913. It ends in 3, so the cube root ends in 7. Leaving out 913 leaves 4, which lies between $1^3 = 1$ and $2^3 = 8$, so the tens digit is 1. The cube root is 17. Check: $17^3 = 4913$.

12167. It ends in 7, so the cube root ends in 3. Leaving out 167 leaves 12, which lies between $2^3 = 8$ and $3^3 = 27$, so the tens digit is 2. The cube root is 23. Check: $23^3 = 12167$.

32768. It ends in 8, so the cube root ends in 2. Leaving out 768 leaves 32, which lies between $3^3 = 27$ and $4^3 = 64$, so the tens digit is 3. The cube root is 32. Check: $32^3 = 32768$.

∴ $\sqrt[3]{1331} = 11$, $\sqrt[3]{4913} = 17$, $\sqrt[3]{12167} = 23$ and $\sqrt[3]{32768} = 32$.

We work out each difference:

$67^3 - 66^3 = 300763 - 287496$
$67^3 - 66^3 = 13267$

$43^3 - 42^3 = 79507 - 74088$
$43^3 - 42^3 = 5419$

$67^2 - 66^2 = 4489 - 4356$
$67^2 - 66^2 = 133$

$43^2 - 42^2 = 1849 - 1764$
$43^2 - 42^2 = 85$

∴ (i) $67^3 - 66^3$ is the greatest.

Reasoning without the full calculation: the difference between the squares of two consecutive numbers is just their sum ($67^2 - 66^2 = 67 + 66 = 133$), but the difference between their cubes is much larger ($67^3 - 66^3 = 67^2 + 67 \times 66 + 66^2$). Cubes grow faster than squares, and bigger numbers give bigger gaps, so the gap between the cubes of the larger numbers, 67 and 66, is the greatest.