3.Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81

(ii) 128

(iii) 135

(iv) 92

(v) 704

(i) We have the prime factors of 81 as:

\(\qquad \begin{array}{c|lcr} 3 & 81\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1 \end{array} \)

\(81 = 3 \times 3 \times 3 \times 3 = 3^3 \times 3\)

We can observe clearly that, number 3 is divided to 81 to make it a perfect cube.

So, 81 ÷ 3 = 27 which is a perfect cube. Thus, the required smallest number to be divided is 3.

(ii) We have the prime factors of 256 as:

\(\qquad \begin{array}{c|lcr} 2 & 128\\ \hline 2 & 64\\ \hline 2 & 32\\ \hline 2 & 16\\ \hline 2 & 8\\ \hline 2 & 4\\ \hline 2 & 2\\ \hline & 1\\ \end{array} \)

\(128 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3 \times 2\)

We can observe clearly that, number 2 is divided to 128 to make it a perfect cube.

So, 128 ÷ 2 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 2.

(iii) We have the prime factors of 135 as:

\(\qquad \begin{array}{c|lcr} 3 & 135\\ \hline 3 & 45\\ \hline 3 & 15\\ \hline 5 & 5\\ \hline & 1 \end{array} \)

\(81 = 3 \times 3 \times 3 \times 5 = 3^3 \times 5\)

We can observe clearly that, number 5 is divided to 135 to make it a perfect cube.

So, 135 ÷ 5= 27 which is a perfect cube. Thus, the required smallest number to be divided is 5.

(iv) We have the prime factors of 192 as:

\(\qquad \begin{array}{c|lcr} 2 & 192\\ \hline 2 & 96\\ \hline 2 & 48\\ \hline 2 & 24\\ \hline 2 & 12\\ \hline 2 & 6\\ \hline 3 & 3\\ \hline & 1\\ \end{array} \)

\(192 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^3 \times 2^3 \times 3\)

We can observe clearly that, number 3 is divided to 192 to make it a perfect cube.

So, 192 ÷ 3 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 3.

(v) We have the prime factors of 704 as:

\(\qquad \begin{array}{c|lcr} 2 & 704\\ \hline 2 & 352\\ \hline 2 & 176\\ \hline 2 & 88\\ \hline 2 & 44\\ \hline 2 & 22\\ \hline 11 & 11\\ \hline & 1\\ \end{array} \)

\(128 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 = 2^3 \times 2^3 \times 11\)

We can observe clearly that, number 11 is divided to 704 to make it a perfect cube.

So, 704 ÷ 11 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 11.