Question

what is the smallest number by which 35721 must be divided so that the quotient is a perfect cube​

Answer 1

Given: the number is 35721

To find: what is the smallest number by which 35721 must be divided so that the quotient is a perfect cube

Solution:

• 35721 = 3 × 11907
• = 3 × 3 × 3969
• = 3 × 3 × 3 × 1323
• = 3 × 3 × 3 × 3 × 441
• = 3 × 3 × 3 × 3 × 3 × 147
• = 3 × 3 × 3 × 3 × 3 × 3 × 49
• = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

We see that, in prime factorization of 35721, three 3s exist twice and 7s exist two times only.

If we divide the number 35721 by two 7s, i.e., 7 × 7 = 49, then the numebr will become a perfect cube.

Answer: 49 is the smallest number by which 35721 must be divided so that the quotient is a perfect cube.

Answer 2

$\huge\purple\star\underline\mathfrak\pink{Solution \: below \: :-}\purple\star$

$\green { \underline \bold{ given : - }} \: \\ \tt : 35721 \\ \tt : \blue { \underline \bold{ to \: find : - }} \: \\ \tt :the \: no.35721 \: is \: not \: a \: perfect \: square . \\ \tt : \green { \underline \bold{ lets \: find \: by \: prime \: factorisation \: method: - }} \: \\ \tt :lcm \: of \: 35721 \: = \: 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7 = {3}^{3} \times {3}^{3} \times {7}^{2} \\ \tt : according \: to \: the \: prime \: factorisation \: method \\ \tt \: here \: one \: digit \: 7 \: \\ \tt should \: be \: mutilplied \: to \: be \: a \: perfect \: square \\ \tt : \red{ \underline \bold\therefore\: { \: It \: is \: proved \: that \: the \: no.35721 \: is \: not \: a \: perfect \: square : - }}$