The volume V (in cubic centimeters) of a given sample of water varies with changing temperature T (in degrees Celsius). For T between 0°C and 30°C, the relation is given almost exactly by the formula V(T)=V_0 [1-(6.427×10^(-5) )T+(8.505×10^(-6) ) T^2-(6.790×10^(-8) ) T^3]

Where V_0 is the volume of water (not ice) sample at 0°C.

Suppose that V_0=10^5 〖cm〗^3. Substitute this value into the formula above and simplify the resulting formula.
Find the volume when T=20°C and V_0=10^5 〖cm〗^3. Include the units in your answer.
Use the first derivative to find the instantaneous rate of change of volume with respect to temperature when T=20°C and V_0=10^5 〖cm〗^3. Include the units in your answer.
Could anyone give me a hand with this problem?
Cheers