Quadratic approximation of $9^{1/3}$ Find a quadratic approximation of the cube root of $9$ by using the equation $9=8(1+\frac 18)$, and estimate the difference between the exact value and the approximation.
How am I supposed to find the quadratic approximation of this formula? If there is no $x$, and the formula for quad approx is $Qa(x) = f(a) + f '(a)(x-a) + f ''(a)(x-a)^2/2$, so does it mean that my quadratic approx will be equal just to $f(a)$?
Could anyone help please, it looks easy, but I can't get it. 

Since $9=8(1+\frac{1}{8})$, 
$\;\;\sqrt[3]9=2\big(1+\frac{1}{8}\big)^{1/3}=2\big(1+\frac{1}{3}(\frac{1}{8})-\frac{1}{9}(\frac{1}{8})^2+\frac{5}{81}(\frac{1}{8})^3+\cdots\big)$
since $\displaystyle(1+x)^{1/3}=1+\frac{1}{3}x+\frac{(\frac{1}{3})(\frac{1}{3}-1)}{2!}x^2+\frac{(\frac{1}{3})(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}+\cdots$ 

Since the series is alternating (starting with the second term) and has terms decreasing in absolute value ,
the error in using this approximation is at most $2(\frac{5}{81})(\frac{1}{8})^3$, the absolute value of the first omitted term.
