A solid, metallic right circular cone of radius 7 cm and height 3 cm is melted into three cubes. If the sides of two cubes are 3 cm and 5 cm, then the side of the third cube will be how much?

A cone is melted and reformed into 3 cubes. The volume of the cone equals the sum of volumes of the three cubes.

Volume of a cone:

$V_{cone} = \frac{1}{3}\pi r^2h$

$V_{cone} = \frac{1}{3} \times \pi \times (7)^2 \times 3$

$V_{cone} = \frac{1}{3} \times \pi \times 49 \times 3$

$V_{cone} = 49\pi \text{ cm}^3$

Using $\pi = \frac{22}{7}$:

$V_{cone} = 49 \times \frac{22}{7}$

$V_{cone} = 7 \times 22$

$V_{cone} = 154 \text{ cm}^3$

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Volume of cube with side 3 cm:

$V_1 = 3^3$

$V_1 = 27 \text{ cm}^3$

Volume of cube with side 5 cm:

$V_2 = 5^3$

$V_2 = 125 \text{ cm}^3$

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The volume of the cone equals the sum of volumes of all three cubes:

$V_{cone} = V_1 + V_2 + V_3$

$154 = 27 + 125 + V_3$

$154 = 152 + V_3$

$V_3 = 2 \text{ cm}^3$

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For the third cube with volume 2 cm³:

$(side)^3 = 2$

$side = \sqrt[3]{2}$

$side = 2^{1/3}$

Therefore, the side of the third cube is $\sqrt[3]{2}$ cm, which is approximately 1.26 cm or $\sqrt{2}$ cm (approximately 1.414 cm).