Use differences to find a pattern in the sequence.
$4,8,19,45,94,174,293$
Assuming that the pattern continues, the eighth term should be

Step 1 ：First, we calculate the differences between consecutive terms in the sequence: $8-4=4$, $19-8=11$, $45-19=26$, $94-45=49$, $174-94=80$, $293-174=119$.

Step 2 ：The differences are not constant, so we calculate the second differences: $11-4=7$, $26-11=15$, $49-26=23$, $80-49=31$, $119-80=39$.

Step 3 ：The second differences are not constant either, so we calculate the third differences: $15-7=8$, $23-15=8$, $31-23=8$, $39-31=8$.

Step 4 ：The third differences are constant, which means the sequence is a cubic sequence. The general form of a cubic sequence is $an^3+bn^2+cn+d$.

Step 5 ：We know that $a$ is one third of the third difference, so $a=\frac{8}{3}$.

Step 6 ：Substitute $n=1$ into the formula, we get $d=4$.

Step 7 ：Substitute $n=2$ into the formula, we get $\frac{8}{3}*2^3+2b+c+4=8$, which simplifies to $\frac{32}{3}+2b+c=8$.

Step 8 ：Substitute $n=3$ into the formula, we get $\frac{8}{3}*3^3+3^2b+3c+4=19$, which simplifies to $\frac{216}{3}+9b+3c=19$.

Step 9 ：Solving the system of equations $\frac{32}{3}+2b+c=8$ and $\frac{216}{3}+9b+3c=19$, we get $b=\frac{1}{2}$ and $c=\frac{1}{2}$.

Step 10 ：So the general formula of the sequence is $\frac{8}{3}n^3+\frac{1}{2}n^2+\frac{1}{2}n+4$.

Step 11 ：Substitute $n=8$ into the formula, we get $\frac{8}{3}*8^3+\frac{1}{2}*8^2+\frac{1}{2}*8+4=\boxed{457}$.