Question

2) Is 266 200 a perfect cube? If not, find the smallest multiple of 26200 which is a perfect cube find the cube root of the new number.​

Answer 1

$Given \: number \: 266200$

$Resolving \: 266200 \: into \:prime \:factors ,\\we \:get$

2| 266200

_________

2| 133100

_________

2| 66550

_________

5| 33275

_________

5| 6655

_________

11| 1331

_________

11| 121

_________

**** 11

266200 = 2 × 2 × 2×5×5×11×11×11

The prime factor 5 does not appear in a group of three.

So, 266200 is not a perfect cube.

Hence, the smallest number by which it is to be multiplied to make it a perfect cube is 5 .

$\red{ Perfect \: cube }\\= 266200 \times 5 \\= 2^{3} \times 5^{3} \times 11^{3} \\= ( 2\times 5 \times 11)^{3} \\= (110)^{3}$

$\red{ Cube \:root \: of \: new \: number } \\= \sqrt[3] { 110^{3}} \\\green {= 110}$

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