Three numbers are to one another 2:3:4. The sum of their cubes is 0.334125. Find the numbers.

Given: 

Three numbers are to one another 2:3:4. The sum of their cubes is 0.334125. 

To find: 

We have to find the numbers.

Solution:

Let the numbers be $2a, 3a$ and $4a$

This implies,

$(2a)^3+ (3a)^3+(4a)^3 = 0.334125$

$8a^3+27a^3+64a^3 = 0.334125$

$99a^3 = 0.334125$

$99a^3 = \frac{334125}{1000000}$

$a^3=\frac{334125}{1000000\times99}$

$a^3= \frac{3375}{1000000}$

$a=\sqrt[3]{\frac{3375}{1000000}}$

$a=\frac{\sqrt[3]{(15\times15\times15)}} {\sqrt[3]{(100\times100\times100)}}$

$a= \frac{15}{100}$

$a= 0.15$

$2a=2(0.15)=0.3$

$3a=3(0.15)=0.45$

$4a=4(0.15)=0.6$

Therefore,

The numbers are $0.3, 0.45$ and $0.6$.

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Updated on: 10-Oct-2022

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