Given that x_n = (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) (2x_n + (N)/(x_n²)).

The clock frequency is $f = 3\, \text{MHz} = 3 \times 10^6\, \text{Hz}$.
The time period of one T-state ($T_{T-state}$) is the reciprocal of the clock frequency.

$$T_{T-state} = \frac{1}{f} = \frac{1}{3 \times 10^6}\, \text{s}$$ $$T_{T-state} \approx 0.333 \times 10^{-6}\, \text{s} = 0.333\, \mu\text{s}$$

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