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(C) Let the edges of the three cubes be $3x, 4x,$ and $5x \, 	ext{cm}$ respectively.The volume of the three cubes is $V = (3x)^3 + (4x)^3 + (5x)^3 =

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$6, 8, 10$

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(C) Let the edges of the three cubes be $3x, 4x,$ and $5x \, 	ext{cm}$ respectively.The volume of the three cubes is $V = (3x)^3 + (4x)^3 + (5x)^3 = 27x^3 + 64x^3 + 125x^3 = 216x^3 \, 	ext{cm}^3$.Let the side of the new single cube be $a$. The volume of this new cube is $a^3 = 216x^3$,which implies $a = 6x$.The diagonal of a cube with side $a$ is given by $d = a\sqrt{3}$.Given that the diagonal is $12\sqrt{3} \, 	ext{cm}$,we have $a\sqrt{3} = 12\sqrt{3}$,so $a = 12 \, 	ext{cm}$.Since $a = 6x$,we have $6x = 12$,which gives $x = 2$.Therefore,the edges of the three cubes are $3(2) = 6 \, 	ext{cm}$,$4(2) = 8 \, 	ext{cm}$,and $5(2) = 10 \, 	ext{cm}$.

Three cubes of a metal whose edges are in the ratio $3:4:5$ are melted and converted into a single cube whose diagonal is $12\sqrt{3} \, \text{cm}$. Find the edges of the three cubes (in $\text{cm}$).

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