3 cubes of a metal whose …

Given, Three cubes of a metal whose edges are in the ratio 3 : 4 : 5{tex}{/tex}.So,we can take the edges of these cubes as  3x, 4x and 5x.
Volume of the cones are, (3x){tex}^3{/tex}, (4x){tex}^3{/tex}, (5x){tex}^3{/tex} i.e  27x{tex}^3{/tex},  64x{tex}^3{/tex},  125x{tex}^3{/tex}.
We know that,
the length of the diagonal of a single cube of side {tex}a = a \sqrt { 3 }{/tex}
The length of the diagonal of a cube is given to be {tex}12 \sqrt { 3 }.{/tex}
So, the edge of a single cube is 12 cm.
Volume of the single cube = (12)³ = 1728 cm³
Since the three cubes are melted and converted into a single cube,
Sum of the volumes of the three cubes = Volume of a single cube
{tex}\Rightarrow{/tex} 27x³ + 64x³ + 125x³ {tex}=1728{/tex}
{tex}\Rightarrow{/tex} 216x³ {tex}= 1728{/tex}
{tex}\Rightarrow{/tex} x³ = 8
{tex}\Rightarrow{/tex} {tex}x = 2 \ cm{/tex}
So, the edges of the cubes are 6 cm, 8 cm and 10 cm.