Negative Exponents and Indices - Class VIII

  If the exponent of a negative integer is odd, then the result is a negative integer.

  If the exponent of a negative integer is even, then the result is a $positive$ integer.

$\dfrac {1}{3^{4}} = 3^{-4}$

$1 \div 7 \div 7 = \dfrac 17 \div 7 = \dfrac 17 \times \dfrac 17$

$\dfrac{1}{x} = x^{-1}$. Index of $x$ is $(-1)$, which is negative.

$\dfrac{1}{x^3}=x^{-3}$ . Index of $x$ is  $(-3)$, which is negative.

$\dfrac { 1 }{ { x }^{ 2 } } $ = ${ x }^{ -2 }$
Thus ${ x }^{ -2 }$ has a negative index
Therefore option $A$ is the correct answer.

In option C, $\dfrac{1}{x^2} = x^{-2}$.  

We see that, $\left (\dfrac{3}{2}\right )^{-3}$=$\left (\dfrac{2}{3}\right)^{3}$

It can be written as $\left (\dfrac {4}{3} - 4\right)^{-1}$ $=$ $\left (\dfrac {-8}{3}\right)^{-1}$ $=$ $-\dfrac {3}{8}$

It can be written as,

$\left (\dfrac{1}{3}\right)^{-3}=3^{3}$

$\dfrac 15 = \dfrac {1}{5^1}=5^{-1}$

$-2^{-3} = \dfrac {1}{-2 \times -2 \times -2} = -\dfrac 18$

We have.

$ \sqrt[3]{a+\sqrt{b}}=7+4\sqrt{3} $

$ {{(a+\sqrt{b})}^{1/3}}=(7+4\sqrt{3}) $

$ a+\sqrt{b}={{(7+4\sqrt{3})}^{3}}\ \ ......\ \ (1) $

$ \text{Similarly,} $

$ a-\sqrt{b}={{(7-4\sqrt{3})}^{3}}\ \ ......\ \ (2) $

On multiplying (1) and (2) to. We get,

$ (a+\sqrt{b})(a-\sqrt{b})={{(7+4\sqrt{3})}^{3}}{{(7-4\sqrt{3})}^{3}} $

$ {{a}^{2}}-b={{\left[ (7+4\sqrt{3})(7-4\sqrt{3}) \right]}^{3}} $

$ {{a}^{2}}-b={{\left[ 49-16\times 3 \right]}^{3}} $

$ {{({{a}^{2}}-b)}^{1/3}}=(1) $

$ \sqrt[3]{{{a}^{2}}-b}=1 $

Hence, this is the
answer

$\displaystyle(x+y)^{-1}(x^{-1}+y^{-1})=\frac{1}{x+y}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x+y}\times\frac{x+y}{xy}=\frac{1}{xy}$

$\displaystyle{1-[1-(1-n)^{-1}]^{-1}}^{-1}=\left[1-\left(1-\frac{1}{1-n}\right)^{-1}\right]^{-1}$

$\displaystyle=\left[1-\left(\frac{1-n-1}{1-n}\right)^{-1}\right]^{-1}=\left[1+\left(\frac{n}{1-n}\right)^{-1}\right]^{-1}=\left(1+\frac{1-n}{n}\right)^{-1}$

$\displaystyle=\left(\frac{n+1-n}{n}\right)^{-1}=\left(\frac{1}{n}\right)^{-1}=n$