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. Text Solution Solution : By using prime factorization we could find:<br>(i)`7803`<br>`675=3^3times17^2`<br>As we can see that `17` is not cubed.<br>Hence, `17` is the smallest number by which `7803` should be multiplied to make it a perfect cube.<br><br>(ii)`107811`<br>`107811=3^3times3times11^3`<br>As we can see that `3` is not cubed.<br>Hence, `3` is the smallest number by which `107811` should be multiplied to make it a perfect cube.<br>(iii)`35721`<br>`35721=3^6times7^2`<br>As we can see that `7` is not cubed.<br>Hence, `7` is the smallest number by which `35721` should be multiplied to make it a perfect cube. |