Question

Is 9720 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

Answer 1

$\begin{array}{cc}2& 9720\\ 2& 4860\\ 2& 2430\\ 3& 1215\\ 3& 405\\ 3& 135\\ 3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}$
Prime factors of 9720 $=2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 5$

(2 marks)

The prime factors 3 and 5 do not appear in group of triplets.
So, 9720 is not a perfect cube.
If we divide the number by $3\times 3\times 5$, then the prime factorisation of the quotient will not contain $3\times 3\times 5=45$

(1 mark)

$\therefore 9720÷45=\underset{–}{2\times 2\times 2}\times \underset{–}{3\times 3\times 3}$
$=216=(6{)}^3$

(1 mark)

Hence, the smallest number by which 9720 should be divided to get a perfect cube, is 45.