Three solid spheres of gold whose radii are $1\text{cm}$, $6\text{cm}$ and $8\text{cm}$ respectively are melted into a single solid sphere. Then the radius of the sphere is:

When solid objects are melted and recast into a new single shape, the total volume remains constant.
The formula for the volume of a sphere is:

$$V=\frac{4}{3}\pi r^3$$

Let the radius of the new single solid sphere be $R$.

$$\text{Volume of new sphere}=\text{Sum of volumes of three individual spheres}$$

$$\frac{4}{3}\pi R^3=\frac{4}{3}\pi r_1^3+\frac{4}{3}\pi r_2^3+\frac{4}{3}\pi r_3^3$$

Factoring out $\frac{4}{3}\pi$ from both sides:

$$R^3=r_1^3+r_2^3+r_3^3$$

Substituting the given radii values ($r_1=1$, $r_2=6$, $r_3=8$):

$$R^3=1^3+6^3+8^3$$

$$R^3=1+216+512$$

$$R^3=729$$

Taking the cube root on both sides:

$$R=\sqrt[3]{729}=9\text{cm}$$

Thus, the radius of the newly formed sphere is $9\text{cm}$.