Gujarat BoardEnglish MediumSTD 8MATHSCubes and Cube Roots2 Marks
Question
Which of the following are perfect cube$?\ 1728$
✓
Answer
On factorising $1728$ into prime factors, we get $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$ Group the factors in triples of equal factors as: $1728 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\}$ It is evident that the prime factors of $1728$ can be grouped into triples of equal factors and no factor is left over. Therefore, $1728$ is a perfect cube.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.