Question

Is 1188 a perfect cube number? If not by which smallest natural number should 1188 be divided so that the quotient is a perfect cube number. Also find the cube root of the new cube number.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{prime}\mathrm{factorisation}\mathrm{of}1188\mathrm{is} \\ 1188=2\times 2\times 3\times 3\times 3\times 11 \\ \mathrm{Since}\mathrm{the}\mathrm{prime}\mathrm{factor}2\mathrm{and}11\mathrm{do}\mathrm{not}\mathrm{occurr}\mathrm{in}\mathrm{the}\mathrm{group}\mathrm{of}3,\mathrm{so}1188\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{perfect}\mathrm{cube}. \\ \mathrm{Now},\mathrm{to}\mathrm{make}1188\mathrm{a}\mathrm{perfect}\mathrm{cube},\mathrm{it}\mathrm{must}\mathrm{be}\mathrm{divided}\mathrm{by}2\times 2\times 11\mathrm{that}\mathrm{is}44. \\ \mathrm{Now},1188÷44=27\left(\mathrm{a}\mathrm{perfect}\mathrm{cube}\mathrm{number}\right) \\ \mathrm{Now},\sqrt[3]{27}=\sqrt[3]{3\times 3\times 3}=3 \end{gathered}$$