Question

Three numbers are in the ratio 3:4:5 . The sum of their cubes is 0.001728.find the number

Answer 1

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Answer 2

Answer:

Required value of the number is $\frac{1}{50}$

Step-by-step explanation:

Given,three numbers are in the ratio 3:4:5.

Let the first number be 3x years and second number be 4x years and third number be 5x years.

It is also given,sum of their cubes is 0.001728.

So according to question,

${(3x)}^{3} + {(4x)}^{3} + {(5x)}^{3} = 0.001728 \\ 27 {x}^{3} + 64 {x}^{3} + 125 {x}^{3} = 0.001728 \\ 216 {x}^{3} = 0.001728 \\ {x}^{3} = \frac{0.001728}{216} \\ {x}^{3} = \frac{1728}{1000000 \times 216} \\ x = \frac{12}{100 \times 6} \\ x = \frac{1}{50}$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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