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# For of the Non-perfect Cubes in Q. No. 20 Find the Smallest Number by Which It Must Be Divided So that the Quotient is a Perfect Cube. - Mathematics

Sum

For of the non-perfect cubes in Q. No. 20 find the smallest number by which it must be divided so that the quotient is a perfect cube.

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#### Solution

The only non-perfect cube in question number 20 is 243.

On factorising 243 into prime factors, we get: $243 = 3 \times 3 \times 3 \times 3 \times 3$ On grouping the factors in triples of equal factors, we get:

$243 = \left\{ 3 \times 3 \times 3 \right\} \times 3 \times 3$It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is divided by ( $3 \times 3 = 9$), the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243 should be divided by 9 to make it a perfect cube.

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#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 4 Cubes and Cube Roots
Exercise 4.1 | Q 21.2 | Page 9
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