Question

Three numbers are in the ratio of $2:3:4$. The sum of their cubes is $33957$. Find the largest number.

Answer 1

Let the number by $2x,3x$ and $4x$.
Then, ${\left(2x\right)}^3+{\left(3x\right)}^3+{\left(4x\right)}^3=33957$
$8x^3+27x^3+64x^3=33957$
$99x^3=33957$
$x^3=\frac{33957}{99}=343$
$x^3=7\times 7\times 7$ [Resolving $343$ into prime factors]
$x=\sqrt[3]{37\times 7\times 7}=7$
Hence, the numbers are
$2x=2\times 7=14$
$3x=3\times 7=21$
and $4x=4\times 7=28$