What is the smallest number by which the following must be multiplied so that the product is a perfect cube 7803?

Question 10 Cube and Cube Roots Exercise 4.1

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Answer:

(i) 675

First find the factors of 675

675 = 3 × 3 × 3 × 5 × 5

= 33 × 52

∴To make a perfect cube we need to multiply the product by 5.

(ii) 1323

First find the factors of 1323

1323 = 3 × 3 × 3 × 7 × 7

= 33 × 72

∴To make a perfect cube we need to multiply the product by 7.

(iii) 2560

First find the factors of 2560

2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

= 23 × 23 × 23 × 5

∴To make a perfect cube we need to multiply the product by 5 × 5 = 25.

(iv) 7803

First find the factors of 7803 7803 = 3 × 3 × 3 × 17 × 17

= 33 × 172

∴To make a perfect cube we need to multiply the product by 17.

(v) 107811

First find the factors of 107811 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

= 33 × 3 × 113

∴To make a perfect cube we need to multiply the product by 3 × 3 = 9.

(vi) 35721

First find the factors of 35721

35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

= 33 × 33 × 72

∴To make a perfect cube we need to multiply the product by 7.

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What smallest number should 7803 be multiplied with so that product becomes a perfect cube

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On factorising 7803 into prime factors, we get:

\[7803 = 3 \times 3 \times 3 \times 17 \times 17\]

On grouping the factors in triples of equal factors, we get:

\[7803 = \left\{ 3 \times 3 \times 3 \right\} \times 17 \times 17\]

It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube. However, if the number is multiplied by 17, the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 7803 should be multiplied by 17 to make it a perfect cube.