1. ## function question

I don't understand why I'm not able to manipulate this quite basic function into another form

the function is:
$\\x=\sqrt[3]{9}\\x^{3}=9\\f(x)=x^{3}-9$

But then this function is exactly the same:
$\\x=\sqrt[3]{9}\\x-\sqrt[3]{9}=0\\\text{If i try to produce a 9 and a}\;x^3\;\text{by cubing both sides}\\(x^3-\sqrt[3]{9})^3=x^3-3x^2\times\sqrt[3]{9}+3x\times(\sqrt[3]{9})^2-9\\\neq x^{3}-9$

Am I misunderstanding something basic here?

2. ## Re: function question

Consider $a^3-b^3 = (a-b)(a^2+ab+b^2)$ so

$x^3-9 = x^3-(\sqrt[3]{9})^3 = (x-\sqrt[3]{9})(x^2 +\sqrt[3]{9}x+(\sqrt[3]{9})^2)$

3. ## Re: Using Newtons method to solve

the actual question was to solve $x=\sqrt[3]{9}$ using Newtons method with the first approximation $x=2$, if you cube both sides and do Newtons method:
$\\x^3=9\\f(x)=x^3-9\\f'(x)=3x^2\\x_2=2\frac{1}{12}=2.0833333333$

but if you continue with the original equation without cubing both sides
$\\f(x)=x-\sqrt[3]{9}\\f'(x)=1\\x_2=\sqrt[3]{9}=2.080083823$

Why are the solutions similar but not exactly the same?

4. ## Re: Using Newtons method to solve

$f(x)=x^3-9$

$\\f(x)=x-\sqrt[3]{9}$

These are different equations, that is why you have different solutions. One is a cubic and one is a straight line.

Newton's will only give an approximation of the root.