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**Hint:**To find the cube root of a number, say p, we have to find the nearest perfect cube, say x. Then, we have to add or subtract y such that the result will be p, that is, $p=x\pm y$ . Now, cube root of p will be equal to $\sqrt[3]{x}\pm \dfrac{y}{3{{\left( \sqrt[3]{x} \right)}^{2}}}$ . Here, we will consider the perfect cube close to 23 as ${{3}^{3}}=27$ . Then, we will write 23 as the difference between 27 and 4. Then, we will substitute these values in the above form.

**Complete step-by-step solution:**

We have to find the cube root of 23. We know that 23 is not a perfect cube. In general, to find the cube root of a number, say p, we have to find the nearest perfect cube, say x.

Now, we have to add or subtract y such that the result will be p.

$\Rightarrow p=x\pm y$

Now, cube root of p can be found as follows:

$\sqrt[3]{x}\pm \dfrac{y}{3{{\left( \sqrt[3]{x} \right)}^{2}}}...\left( i \right)$

So firstly, we have to check for a perfect cube nearest to 23. We know that 23 lies between ${{2}^{3}}=8$ and ${{3}^{3}}=27$ . 27 is closer to 23. Therefore, we will consider 3.

We can write 23 as the difference between 27 and 4.

$\Rightarrow 23=27-4$

Now, we have to write in the form (i).

$\Rightarrow \sqrt[3]{27}-\dfrac{4}{3{{\left( \sqrt[3]{27} \right)}^{2}}}$

We know that ${{3}^{3}}=27$ . Then, the above result becomes

$\begin{align}

& \Rightarrow 3-\dfrac{4}{3\times {{3}^{2}}} \\

& \Rightarrow 3-\dfrac{4}{27} \\

\end{align}$

Let us take the LCM and simplify.

$\begin{align}

& \Rightarrow \dfrac{3\times 27}{1\times 27}-\dfrac{4}{27} \\

& \Rightarrow \dfrac{81-4}{27} \\

\end{align}$

Let us subtract 4 from 81.

$\Rightarrow \dfrac{77}{27}$

We have to divide 77 by 27.

$\Rightarrow \dfrac{77}{27}=2.8$

**Therefore, the cube root of 23 is 2.8.**

**Note:**Students must look for a perfect cube that is very close to the number whose cube root is to be found. We can also find the cube root of 23 using Halley’s method.

Halley’s method is given as follows:

$\sqrt[3]{a}=x\left( \dfrac{{{x}^{3}}+2a}{2{{x}^{3}}+a} \right)$

where, a is the number whose cube root is being calculated and x is integer guess of its cube root that is less than a.

Here, $a=23$ . We know that ${{2}^{3}}=8$ which is less than 23. Therefore, $x=2$ .

Now, let us substitute the values in Halley’s formula.

$\begin{align}

& \Rightarrow \sqrt[3]{23}=2\times \left( \dfrac{{{2}^{3}}+2\times 23}{2\times {{2}^{3}}+23} \right) \\

& \Rightarrow \sqrt[3]{23}=2\times \left( \dfrac{8+46}{16+23} \right) \\

& \Rightarrow \sqrt[3]{23}=2\times \left( \dfrac{54}{39} \right) \\

& \Rightarrow \sqrt[3]{23}=2.76 \\

\end{align}$

We can approximate 2.77 as 2.8.

Therefore, the cube root of 23 is 2.8.

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