Question

4) Three numbers are in the ratio 1 : 2 : 3. If the sum of their cubes is 972, find the numbers.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{three}\mathrm{numbers}\mathrm{be}\mathrm{x},2\mathrm{x}\mathrm{and}3\mathrm{x} \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{question}: \\ {\mathrm{x}}^3+{\left(2\mathrm{x}\right)}^3+{\left(3\mathrm{x}\right)}^3=972 \\ \Rightarrow {\mathrm{x}}^3+8{\mathrm{x}}^3+27{\mathrm{x}}^3=972 \\ \Rightarrow 36{\mathrm{x}}^3=972 \\ \Rightarrow {\mathrm{x}}^3=27 \\ \Rightarrow \mathrm{x}=3 \\ \mathrm{So},\mathrm{the}\mathrm{numbers}\mathrm{are}: \\ 3,6\mathrm{and}9. \end{gathered}$$