Question

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:

1. 243
2. 256
3.  72
4.  675
5.  100

Answer 1

(i) \(243\)

Prime factors of \( 243 =3\times3\times3\times3\times3\)
Here \(3\) does not appear in \(3\)’s group.
Therefore, \(243 \) must be multiplied by \(3\) to make it a perfect cube.

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(ii) \(256\)

Prime factors of \(256\) = \(2\times2\times2\times2\times2\times2\times2\times2\)
Here one factor \(2\) is required to make a \(3\)’s group.
Therefore, \(256\) must be multiplied by \(2\) to make it a perfect cube.

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(iii) \(72\)

Prime factors of \(72\) = \(2 \times 2 \times 2 \times 3 \times 3\)
Here \(3\) does not appear in \( 3\)’s group.
Therefore, \(72\) must be multiplied by \(3\) to make it a perfect cube.

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(iv) \( 675\)

Prime factors of \(675 = 3 \times 3 \times 3\times 5 \times 5\)
Here factor \(5\) does not appear in \(3\)’s group.
Therefore \(675\) must be multiplied by \(5\) to make it a perfect cube.

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(v) \(100\)

Prime factors of \(100 = 2 \times 2 \times 5 \times 5\)
Here factor \(2\) and \(5\) both do not appear in \(3\)’s group.
Therefore \(100\) must be multiplied by \(2 \times5 \) = \(10\) to make it a perfect cube.