Question

three numbers are in the ratio 2 ratio 3 ratio 4 the sum of their cube is 33957 find the number

Answer 1

Let numbers are
2x, 3x, 4x


Given that the sum of their cubes = 33957


(2x)³ + (3x)³ + (4x)³ = 33957

8x³ + 27x³ + 64x³ = 33957

99x³ = 33957

${x}^{3} = \frac{33957}{99}$

x³ = 343

$x = \sqrt[3]{343}$

x = 7



Then, Numbers are :

2x = 2(7) = 14
3x = 3(7) = 21
4x = 4(7) = 28




I hope this will help you


(-:

Answer 2

$\huge{\underline{\bigstar{\sf{solution!!}}}}$

The 3 numbers are : 0.3,0.45,0.6

⇛The question says there are three numbers but with a specific ration what that means in that once we pick one of the numbers the other two are know to us through the rations we can therefore replace all 3 of the numbers with a single variable:

2: 3: 4 ⇛ 2x × 3x × 4x

⇒now, no Matter what we chose for c we get the three numbers into the ratios specified we are also told the sum of the cubes of these three numbers which can write :

$({2x})^{3} + ({3x})^{3} + ({4x})^{3} = 0.334125$

⇛Disturbing the powers across the fact is using

$({a×b})^{c} = {a}^{c} {b}^{c} we\: get :$

${8x}^{3}+ {27x}^{3} + {64x}^{3} = {99x}^{3} = 0.334125$

${x}^{3} = \frac{0.334125}{99} = 0.003375$

${x}^{3} = \sqrt[3]{0.003375} = 0.15$

⇛So the 3 numbers are

$2×0.15, 3×0.15, 4×0.15$

$0.3,0.45,0.6$