How many terms of the series $1^3+2^3+3^3+\dots$ should be taken to get the sum $14400$ ?

Find the sum of the series $\left(2^3-1\right)+\left(4^3-3^3\right)+\left(6^3-{15}^3\right)+\dots$ to $n$ terms.

Find the sum of the series $\left(2^3-1\right)+\left(4^3-3^3\right)+\left(6^3-5^3\right)+\dots$ to $8$ terms.