Three numbers are in the ratio 2 : 3 : 1. The sum of their cubes is 288. Find the numbers.

Three numbers are in the ratio 2 : 3 : 1. So the three numbers are $2x,3x$ and $1x$. Hence,

$(2x)^3 + (3x)^3 + (1x)^3 = 288\\[1em] ⇒ 8x^3 + 27x^3 + 1x^3 = 288\\[1em] ⇒ 36x^3 = 288\\[1em] ⇒ x^3 = \dfrac{288}{36}\\[1em] ⇒ x^3 = 8\\[1em] ⇒ x = \sqrt[3]{8}\\[1em] ⇒ x = \sqrt[3]{2 \times 2 \times 2}\\[1em] ⇒ x = 2\\[1em]$

2x = 2 x 2 = 4

3x = 3 x 2 = 6

Hence, the three numbers are 4, 6 and 2