Three cubes of a metal whose edges are in the ratios 3 : 4 : 5 are melted and converted into a single cube whose diagonal is \[12\sqrt{3}\]. Find the edges of the three cubes.

#### Solution

The three cubes of metal are in the ratio 3 : 4 : 5.

Let the edges of the cubes be 3x, 4x and 5x.

Volume of the three cubes will be

\[V_1 = \left( 3x \right)^3 \]

\[ V_2 = \left( 4x \right)^3 \]

\[ V_3 = \left( 5x \right)^3\]

Diagonal of the single cube = \[12\sqrt{3} cm\]

We know diagonal of the cube = \[a\sqrt{3} = 12\sqrt{3}\]

Hence, the side of the cube = 12 cm

Volume of the bigger cube \[V_b = \left( 12 \right)^3\]

Volume of the three cubes = Volume of the single

\[\left( 3x \right)^3 + \left( 4x \right)^3 + \left( 5x \right)^3 = \left( 12 \right)^3 \]

\[ \Rightarrow 27 x^3 + 64 x^3 + 125 x^3 = 1728\]

\[ \Rightarrow 216 x^3 = 1728\]

\[ \Rightarrow x^3 = \frac{1728}{216} = 8\]

\[ \Rightarrow x = 2\]

Hence, the edges of the three cubes will be \[3 \times \left( 2 \right), 4 \times \left( 2 \right), 5 \times \left( 2 \right) = 6, 8, 10\] cm.