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Answer :
(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.