Answer

Verified

416.4k+ views

**Hint:**We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities to find the simplified form of $-{{216}^{\dfrac{1}{3}}}$ with positive exponents.

**Complete step by step solution:**

We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$. In case the value of $n$ becomes negative, the value of the exponent takes its inverse value.

The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.

If we take two exponential expressions where the exponents are $m$ and $n$.

Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.

The indices get added. So, ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$.

The division works in an almost similar way. The indices get subtracted.

So,

$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.

We also have the identity of ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$.

For given expression $-{{216}^{\dfrac{1}{3}}}$, we find the value of ${{216}^{\dfrac{1}{3}}}$.

For our given expression ${{216}^{\dfrac{1}{3}}}$, we find the prime factorisation of 216.

$\begin{align}

& 2\left| \!{\underline {\,

216 \,}} \right. \\

& 2\left| \!{\underline {\,

108 \,}} \right. \\

& 2\left| \!{\underline {\,

54 \,}} \right. \\

& 3\left| \!{\underline {\,

27 \,}} \right. \\

& 3\left| \!{\underline {\,

9 \,}} \right. \\

& 3\left| \!{\underline {\,

3 \,}} \right. \\

& 1\left| \!{\underline {\,

1 \,}} \right. \\

\end{align}$

Therefore, $216={{2}^{3}}\times {{3}^{3}}$. Taking cube root and applying ${{a}^{mn}}={{\left(

{{a}^{m}} \right)}^{n}}$, we get

${{216}^{\dfrac{1}{3}}}={{\left( {{2}^{3}}\times {{3}^{3}} \right)}^{\dfrac{1}{3}}}={{2}^{3\times

\dfrac{1}{3}}}\times {{3}^{3\times \dfrac{1}{3}}}=6$. We can also express as

${{216}^{\dfrac{1}{3}}}={{\left( {{6}^{3}} \right)}^{\dfrac{1}{3}}}={{6}^{3\times \dfrac{1}{3}}}=6$.

So, $-{{216}^{\dfrac{1}{3}}}=-6$

**Therefore, the simplified form of $-{{216}^{\dfrac{1}{3}}}$ is $-6$.**

**Note:**The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.

For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$. We need to remember that the condition for ${{a}^{m}}={{a}^{n}}\Rightarrow m=n$ is that the value of $a\ne 0,\pm 1$.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

A rainbow has circular shape because A The earth is class 11 physics CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell