x^3 - 70 = 0
Now we add 70 to both sides to single out x.
x^3 = 70
Now we take the cube root of both sides:
x = cubert(70)
Thus x = the cube root of 70.
Now the Newton Raphson Method states this:
X_n+1 = X_n - f(X_n)/f'(X_n)
Where X_n+1 = first approximation and X_n is the initial estimate and f'(X_n) is the derivative of the function at the initial estimate. Since the initial estimate is 4 we just plug in and solve.
X_n+1 = 4 - (4^3 - 70)/(3*4^2)
The derivative of x^3 -70 is 3x^2 so 3(4)^2 checks out.
X_n+1 = 4.125
The cube root of 70 = 4.1213
So they are close.
If we keep using the formula, the approximations are as follows:
And there you go. Two consecutive approximations that are exact to 3 decimal places. If you would like to calculate and verify you may. This also seems to converge right onto the square root of 70, which is the root of the original function f(x) = x^3 - 70
I hope this helps.