Given the mean & variance of a dataset of linear measurements X ... N~(14,2), I want to find the mean of X^3.
So by taking mean as being expected value, & rearranging: -
$$Var(X)=E[X^2]−(E[X])^2$$
I get
$$ E[X^2] = Var(X) + (E[X])^2$$
$$E[X^2] = SD^2 + (E[X])^2$$
How can I go from this to using an exponent of 3 to solve the original problem?
Easy way is to use moment generating functions. For the normal this is
$$M(t) = e^{t\mu + \sigma^2t^2/2}$$
Then we can get the moments by using
$$\text{E}[X^n] = \left.\dfrac{d^n M(t)}{dt^n}\right|_{t=0}$$
the $n$th moment is the $n$th derivative of the moment generating function evaluated at 0. For your particular example we get
$$\begin{align*}M(t) &= e^{t(14) + (2)t^2/2}\\ \dfrac{d^3 M(t)}{dt^3} &= (8t^3 + 168t^2 + 1188t +2828)e^{t^2+14t}\\ \left.\dfrac{d^3 M(t)}{dt^3}\right|_{t=0} &= 2828 \end{align*}$$
