Although this may not be the required method, it is a solution, so once Bairstow’s Method has been applied and a solution obtained, it should match the solution given below (part of which includes Newton’s Method, which is in fact incorporated into Bairstow’s Method).

First thing to note is that this is a cubic and it has two complex roots and one real root at x=3 (approx).

So we can use the iterative formula x=x-f(x)/f'(x) where the LHS x is the next iteration of the RHS x.

x₁=x₀-(x³-4x²+3x-1)/(3x²-8x+3).

Therefore, plugging in x₀=3:

x₁=19/6,

x₂=85/27,

x₃=3.14789908,

x₄=3.147899036,

x₅=3.1478990357.

This is a root of the cubic.

By dividing by it we will obtain a quadratic.

Using synthetic division:

3.1478990357 | 1 -4 3 | -1

__1 3.1478990357 -2.682327804 | 1__

1 -0.8521009643 0.3176721962 | 0

Quadratic:

x²-0.8521009643x+0.3176721962.

Therefore root x=(0.8521009643±√(0.8521009643²-4×0.3176721962))/2.

x=0.4260504821±0.3689894075i.

If we take this to 5 decimal places we get the total solution:

x=3.14790, 0.42605+0.36899i, 0.42605-0.36899i.