Suppose that a^2 varies inversely with b^3. If a=3 when b=7 find the value of a^2 when b=6.
a^2 = k / b^3 where k is the proportionality constant
3^2 = k / (7)^3
k = 3^2 / 7^3 = 9 / 343
So
a^2 = (9/343) / 6^3 = 9 / [343 * 6^3 ] = 9 / 74088 = 1/ 8232