Given root of polynomial function x=∛28

g(2)=x^{3} -28=0

Applying Newton's method

x_(n+1) = x_n - (g(x_n)) / (g'(x_n))

For c₁ ⇒ c₁ = c₀ - (g(c₀)) / (g'(c₀))

Thus, c₀ =3

So, g'(3)=27

g(3)=27-28=-1

c₁ = (3-(-1)) / (27) = (81+1) / (27)

c₁ = (82)/(27)

Given root of polynomial function x=∛28

g(2)=x^{3} -28=0

Applying Newton's method

x_(n+1) = x_n - (g(x_n)) / (g'(x_n))

For c₁ ⇒ c₁ = c₀ - (g(c₀)) / (g'(c₀))

Thus, c₀ =3

So, g'(3)=27

g(3)=27-28=-1

c₁ = (3-(-1)) / (27) = (81+1) / (27)

c₁ = (82)/(27)

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