1.Which of the following numbers are not perfect cubes?
(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

(i) We have the prime factors of 216 as:

\( \begin{array}{c|lcr} 2 & 216\\ \hline 2 & 108\\ \hline 2 & 54\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1 \end{array} \)

\(216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\)

Thus, 216 is a perfect cube because in the factorisation, 2 and 3 have formed a group of three.

(ii) We have the prime factors of 128 as:

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

\(128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\)

Here, 2 is left without making a group of three.So, 128 is not a perfect cube.

(iii) We have the prime factors of 1000 as:

\( \begin{array}{c|lcr} 2 & 1000\\ \hline 2 & 500\\ \hline 2 & 250\\ \hline 5 & 125\\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1 \end{array} \)

\(1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5\)

Thus, 1000 is a perfect cube because all the factors make perfect pairs.

(iv) We have the prime factors of 100 as:

\( \begin{array}{c|lcr} 2 & 100\\ \hline 2 & 50\\ \hline 5 & 25\\ \hline 5& 5\\ \hline & 1 \end{array} \)

\(100 = 2 \times 2 \times 5 \times 5\)

Thus, 100 is not a perfect cube because 2 and 5 have not formed a required pair of three.

(v) We have the prime factors of 46656 as:

\( \begin{array}{c|lcr} 2 & 46656\\ \hline 2 & 23328\\ \hline 2 & 11664\\ \hline 2 & 5832\\ \hline 2 & 2916\\ \hline 2 & 1458\\ \hline 3 & 729\\ \hline 3 & 243\\ \hline 3 & 81\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1 \end{array} \)

\(46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3\)

Thus, 46656 is a perfect cube because all the factors are in perfect pairs of three.