You May Need Calculus

Let $\sqrt[3]{27.2} = 3+\delta$ . Then, we have:

$\begin{aligned} (3+\delta)^3 & = 27.2 \\ 27+27\delta+9\delta^2 + \delta^3 & = 27.2 \\ 27\delta+9\delta^2 + \delta^3 & = 0.2 & \small \color{#3D99F6} \text{Since }\delta \text{ is small} \\ 27 \delta & \approx 0.2 & \small \color{#3D99F6} \text{First approximation} \\ \implies \delta_1 & \approx 0.0074 \\ 9\delta^2 + 27 \delta & \approx 0.2 & \small \color{#3D99F6} \text{Second approximation} \\ 9\delta_2^2 + 27 \delta_2 - 0.2 & = 0 \\ \implies \delta_2 & \approx 0.0074 \end{aligned}$

$\implies \sqrt[3]{27.2} \approx \boxed{3.007}$

To solve this, I have found that a very good method to approximate the root of equations is to use a Taylor series, since we can convert the function into a polynomial, which we can evaluate in an easier fashion:

We can first define a Taylor series as:

$g(x) \ = \ \displaystyle\sum_{n=0}^{\infty} \ f^{(n)}(x_0) \ \dfrac{(x \ - \ x_0)^n}{n!}$

Where for a finite number of terms:

$g(x) \ \approx \ f(x)$

We can then see to solve this problem, we can set $f(x) \ = \ \sqrt[3]{x}$ . Next, we can start to solve this. We can probably solve $3$ terms in this Taylor series to get an approximation to $3$ decimal places, so we must do that. We can evaluate the these terms term as $n \ = \ 0, 1, 2$ . We have:

$g(x) \ = \ \displaystyle\sum_{n=0}^{\infty} \ f^{(n)}(x_0) \ \dfrac{(x \ - \ x_0)^n}{n!}$

So we can substitute in our values. We can also centre this function at $27$ , since it is the closest perfect cubic root, therefore $x_0 \ = \ 27$ . We have:

$g(x) \ = f^{(0)}(x_0) \ \dfrac{(x \ - \ x_0)^0}{0!} \ + \ f^{(1)}(x_0) \ \dfrac{(x \ - \ x_0)^1}{1!} \ + \ f^{(2)}(x_0) \ \dfrac{(x \ - \ x_0)^2}{2!}$

We can then simplify. For all of the $f^n$ s, I am taking the derivative, specified by what $n$ is equal to.

$g(x) \ = 3 \ + \ \dfrac{1}{3x_0^{\frac{2}{3}}} \ (x \ - \ x_0) \ - \ \dfrac{2}{9x_0^{\frac{5}{3}}} \ \dfrac{(x \ - \ x_0)^2}{2}$

We can then substitute in $27$ for all of the $x_0$ values in the equation:

$g(x) \ = 3 \ + \ \dfrac{1}{3(27)^{\frac{2}{3}}} \ (x \ - \ 27) \ - \ \dfrac{2}{9(27)^{\frac{5}{3}}} \ \dfrac{(x \ - \ 27)^2}{2}$

We can then evaluate:

$g(x) \ = 3 \ + \ \dfrac{1}{27} \ (x \ - \ 27) \ - \ \dfrac{2}{2187} \ \dfrac{(x \ - \ 27)^2}{2}$

We can then substitute in $27.2$ for all of the $x$ values in the equation:

$g(x) \ = 3 \ + \ \dfrac{1}{27} \ (27.2 \ - \ 27) \ - \ \dfrac{2}{2187} \ \dfrac{(27.2 \ - \ 27)^2}{2}$

$g(x) \ = 3 \ + \ \dfrac{1}{27} \ (0.2) \ - \ \dfrac{2}{2187} \ \dfrac{(0.2)^2}{2}$

We know that $0.2 \ = \ \dfrac{1}{5}$ , so:

$g(x) \ = 3 \ + \ \dfrac{1}{27} \ (\frac{1}{5}) \ - \ \dfrac{2}{2187} \ \dfrac{(\frac{1}{5})^2}{2}$

Then:

$g(x) \ = 3 \ + \ \dfrac{1}{27} \ (\frac{1}{5}) \ - \ \dfrac{2}{2187} \ \dfrac{1}{50}$

$g(x) \ = 3 \ + \ \dfrac{1}{135} \ - \ \dfrac{2}{109350}$

Finally, we can get that:

$g(x) \ \approx \ 3.007$

Therefore:

$\sqrt[3]{27.2} \ \approx \ \boxed{3.007}$

$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

If $f(x)=\sqrt[3]{x}$ , $x=27$ and $h=0.2$ , we can find approximate value of $\sqrt[3]{27.2}$

$f'(x)=\frac{1}{3x^\frac{2}{3}}$

Thus, $\frac{1}{3*9}=\frac{f(x+h)-3}{0.2}$

$\frac{1}{3*9*5}=f(x+h)-3$

$\frac{1}{135}=f(x+h)-3$

$f(x+h)=3+0.0074$

$f(x+h) \approx 3.007$

$\text{Let the equation be, }y = \sqrt[3]{x} = x^{\dfrac{1}{3}} \\ y' = \dfrac{1}{3}\sqrt[3]{x^{-2}} = \dfrac{x^{-\dfrac{2}{3}}}{3} \\ \text{Let's try to approximate the answer using known cubes, } 3^3 = 27 \\ \text{So, the slope is} \\ x = 27 \implies \dfrac{1}{3} \times 27^{-\dfrac{2}{3}} \\ = \dfrac{1}{27} \\ \text{Therefore, we get a linear approximation} \\ y = 3 + \dfrac{1}{27} \left(x - 27\right)\\ y = 3 + \dfrac{x}{27} - 1 \\ y = 2 + \dfrac{x}{27} \\ \text{Now, for x = 27.2} \\ y = 2 + \dfrac{27.2}{27} = 2 + \dfrac{27 + 0.2}{27} \\ = 3 + \dfrac{2}{270} \\ \boxed{y \approx 3.0074074074 \approx 3.007}$