find the least number that should be multiplied by 54900 to make it a perfect cube. also find the cube root of the product .

Dear Student,

Please find below the solution to the asked query:
$$\begin{gathered} \mathrm{First} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{factorise} 54900 \\ 54900=2\times 2\times 5\times 5\times 3\times 3\times 61 \\ \mathrm{For} \mathrm{perfect} \mathrm{cube} \mathrm{make} \mathrm{all} \mathrm{the} \mathrm{factors} \mathrm{in} \mathrm{three} \mathrm{pairs} \\ \mathrm{So} \mathrm{for} \mathrm{that} \mathrm{we} \mathrm{need} \mathrm{one} 2 , \mathrm{one} 5 , \mathrm{one} 3 , \mathrm{two} 61 \\ \mathrm{So} 2\times 5\times 3\times 61\times 61=30\times 61\times 61=111630 \\ \mathrm{Hence} 111630 \mathrm{is} \mathrm{the} \mathrm{least} \mathrm{number} \left(\mathrm{Answer}\right) \\ \mathrm{Cube} \mathrm{root} \mathrm{of} \mathrm{product}=54900\times 111630 \\ \mathrm{Use} \mathrm{calculator} \mathrm{it} \mathrm{will} \mathrm{be} =1830 \\ \mathrm{Or} \mathrm{As} \mathrm{all} \mathrm{the} \mathrm{pairs} \mathrm{are} \mathrm{in} \mathrm{cube} \mathrm{pair} \mathrm{form} \mathrm{take} \mathrm{one} \mathrm{and} \mathrm{multiply} \mathrm{each} \mathrm{other} \\ \mathrm{So} \\ \Rightarrow 2\times 3\times 5\times 61=30\times 61=1830 \left(\mathrm{Answer}\right) \end{gathered}$$

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