Question

three numbers are in the ratio of 2:3:4. the sum of their cubes is 33957. Find the numbers

Answer 1

x = 14
y = 21
z = 28

Answer 2

Method 1:
Let k be a common integer multiplier. Then\[(2k)/(3k)/(4k)=2/3/4\]\[(2k)^3+(3k)^3+(4k)^3\text{=}33957\]\[99 k^3\text{= }33957\]\[k^3\text{=}343\]\[k=\sqrt[3]{343}=7\]\[(2*7)^3+(3*7)^3+(4*7)^3=(14)^3+(21)^3+(28)^3=33957 \]
Method 2:\[\text{Table}\left[\left\{(2k)^3+(3k)^3+(4k)^3,\{2k,3k,4k\}\right\},\{k,6,8\}\right] \]\[\{\{21384,\{12,18,24\}\},\{33957,\{14,21,28\}\},\{50688,\{16,24,32\}\}\} \]The first line is a small Mathematica program which evaluates the LHS of the equation in Method 1, calculates the three numbers in the ratio of 2:3:4 to be cubed and then makes a list of the results where k is ranged from 6 through 8 respectively. The central list of output results is the answer to the problem.