Question

Three no.s are in the ratio 3:4:5 and the sum their cubes is 1728. Find the numbers.

Answer 1

Given :

• Three numbers are in the ratio 3:4:5

• Sum of their cubes is 1728

To Find :

• The three numbers

Solution :

Let the numbers be 3x, 4x and 5x

$\bf \underline{According \: to \: the \: Question : }$

$\longrightarrow \sf \: ({3x})^{3} + (4x) ^{3} + ( {5x})^{3} = 1728$

$\longrightarrow \sf \: 27 {x}^{3} + 64x ^{3} + 125 {x}^{3} = 1728$

$\longrightarrow \sf \: 216 {x}^{3} = 1728$

$\longrightarrow \sf \: {x}^{3} = \cancel\dfrac{1728}{216}$

$\longrightarrow \sf \: {x}^{3} = 8$

$\longrightarrow \sf \: {x}= \sqrt[3]{8}$

$\longrightarrow \sf \: {x}= \sqrt[3]{2 \times 2 \times 2}$

$\longrightarrow \sf \red {x \: = 2}$

Now, We've assumed the number as 3x,4x and 5x

Numbers :

$\dag \: \sf {3x = 3(2) = \boxed{ \purple{\sf \: 6}}}$

$\dag \: \sf {4x = 4(2) = \boxed{ \purple{\sf \: 8}}}$

$\dag \: \sf {5x = 5(2) = \boxed{ \purple{\sf \: 10}}}$

Verification :

$\longrightarrow \sf \: ({3x})^{3} + (4x) ^{3} + ( {5x})^{3} = 1728$

$\longrightarrow \sf \: ({6})^{3} + (8) ^{3} + ( {10})^{3} = 1728$

$\longrightarrow \sf \: 216 + 512 + 1000 = 1728$

$\longrightarrow \sf 1728= 1728$

Hence, Verified!

Answer 2

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$

In Attatched File.....