Question

Is 1188 a perfect cube? If not, by which smallest natural number should
1188 be divided so that the quotient is a perfect cube?

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Answer 1

Answer

$1188 = 2 \times 2 \times 3 \times 3 \times 3 \times 11$

The primes 2 and 11 do not appear in groups of three. So, 1188 is not a perfect cube. In the factorisation of 1188 the prime 2 appears only two times and the prime 11 appears once. So, if we divide 1188 by 2×2×11=44

Then the prime factorisation of the perfect cube is 44

$resulting \: perfect \: cube \: is \: 1188 \div 44 = 27( = 3) {}^{3}$

Hence proved!!

Answer 2

Prime factorisation of 1188:

$1188 = {2}^{2} \times {3}^{3} \times 11$

Since the factors of 1188 are not in powers of 3 or its multiples, 1188 is not a perfect cube.

So, we need to divide 1188 by a number such that the quotient is having factors with powers of 3 or its multiples.

For this we need to eliminate 2² * 11 from 1188,i.e., divide it by:

2² * 11 = 4 * 11 = 44.

$\frac{1188}{44} = \frac{594}{22} = \frac{297}{11} = \frac{27}{1} = 27$

Checking if 27 is a perfect cube:

Prime factorisation of 27:

$27 = {3}^{3}$

So, 27 is a perfect cube.

Therefore, 44 is the smallest number by which 1188 should be divided so that the quotient is a perfect cube.