Question

The multiplicative inverse of ${{\left( \dfrac{1}{2} \right)}^{4}}\div {{\left( \dfrac{1}{3} \right)}^{4}}+{{\left( -\dfrac{1}{2} \right)}^{3}}$ is

Answer 1

Hint: We solve this question by expanding each of the terms one by one according to their power and then we perform the operations between the terms given which are division and addition. We then get an answer in the form of a fraction $\dfrac{a}{b}.$ The multiplicative inverse of this is nothing but the reciprocal of this given as $\dfrac{b}{a}.$

Complete step by step answer:
In order to solve this question, let us simplify the given expression first by expanding the powers for each individual term.
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{4}}\div {{\left( \dfrac{1}{3} \right)}^{4}}+{{\left( -\dfrac{1}{2} \right)}^{3}}$
We know that a term x raised to the power n is nothing but the term x multiplied n times. This is given as ${{x}^{n}}=x\times x\times x\times x\times \ldots n\text{ times}\text{.}$ Using this for the above equation,
$\Rightarrow \left( \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2} \right)\div \left( \dfrac{1}{3}\times \dfrac{1}{3}\times \dfrac{1}{3}\times \dfrac{1}{3} \right)+\left( -\dfrac{1}{2}\times -\dfrac{1}{2}\times -\dfrac{1}{2} \right)$
Since there are no common factors between the numerator and denominator, we multiply all the numerator and denominators and write each individual term.
$\Rightarrow \left( \dfrac{1}{16} \right)\div \left( \dfrac{1}{81} \right)+\left( -\dfrac{1}{8} \right)$
We have got a negative third term since we are multiplying an odd number of negative terms which will give us a negative number. Now, we convert the division into multiplication by taking a reciprocal of the second term and we multiply the negative sign of the third term with the plus sign outside,
$\Rightarrow \left( \dfrac{1}{16} \right)\times \left( \dfrac{81}{1} \right)-\dfrac{1}{8}$
Multiplying the first two terms,
$\Rightarrow \dfrac{81}{16}-\dfrac{1}{8}$
Taking an LCM for the denominator terms 16 and 8, we get the LCM as 16 itself.
$\Rightarrow \dfrac{81\times 1}{16\times 1}-\dfrac{1\times 2}{8\times 2}$
Multiplying the terms,
$\Rightarrow \dfrac{81}{16}-\dfrac{2}{16}$
Since the denominators are same, we subtract the two numerators as
$\Rightarrow \dfrac{81-2}{16}=\dfrac{79}{16}$
Therefore, simplifying the above expression gives us $\dfrac{79}{16}$ and its multiplicative inverse is given as $\dfrac{16}{79}.$

Note: We need to be careful while multiplying or taking the power of a negative number. Even powers of a negative number give us a positive number, odd powers of a negative number give us a negative number.