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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5
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.