Given that xn = (1)/([3]N) is an approximation of the cube root of N. Use Newton-Raphson method to show that a better approximation is given by: xn+1 = (1)/(3) (2xn + (N)/(xn2)).

Step 1: Calculate the time duration of one T-state. The clock frequency is $f = 3\, MHz = 3 \times 10^6\, \text{Hz}$. The time period of one T-state ($T_{T-state}$) is the reciprocal of the clock frequency. $T_{T-state} = \frac{1}{f} = \frac{1}{3 \times 10^6}\, \text{s}$ $T_{T-state} \approx 0.333 \times 10^{-6}\, s = 0.333\, \mu\text{s}$

Step 2: Calculate the total T-states for the delay loop. The delay loop consists of the instructions from AGAIN: to JNZ AGAIN. The T-states for the instructions (assuming an 8085 microprocessor) are: • ADD C: 4 T-states • DCR C: 4 T-states • JNZ AGAIN: 10 T-states (when the jump occurs, i.e., C is not zero) • JNZ AGAIN: 7 T-states (when the jump does not occur, i.e., C is zero)

Let $N$ be the initial value loaded into