A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is

Let the original cone have height $H=27\text{cm}$ and volume $V$.

The cone is cut along a plane parallel to its base at a height of 18 cm from the base. This means the smaller cone formed at the top has a height of:
$h=27-18=9\text{cm}$

Since the smaller cone is similar to the original cone, the ratio of their heights is:
$\frac{h}{H}=\frac{9}{27}=\frac{1}{3}$

The ratio of the volume of the smaller cone ($V_1$) to the volume of the original cone ($V$) is the cube of the ratio of their heights:
$\frac{V_1}{V}={\left(\frac{1}{3}\right)}^3=\frac{1}{27}$
Thus, $V_1=\frac{V}{27}$.

The volume of the second piece (the frustum, $V_2$) is:
$V_2=V-V_1=V-\frac{V}{27}=\frac{26}{27}V$.

The difference in volume between the two pieces is given as 225 cc:
$V_2-V_1=225$
$\frac{26}{27}V-\frac{1}{27}V=225$
$\frac{25}{27}V=225$
$V=225\times \frac{27}{25}$
$V=9\times 27=243\text{cc}$.