Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}a=3^{-3}-3^3\mathrm{and}b=3^3-3^{-3},\mathrm{then}\frac{a}{b}-\frac{b}{a}= \\ \left(\mathrm{a}\right)0\left(\mathrm{b}\right)1\left(\mathrm{c}\right)-1\left(\mathrm{d}\right)2 \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Since},a=3^{-3}-3^3 \\ =\frac{1}{3^3}-3^3\left(\mathrm{As},x^{-m}=\frac{1}{x^m}\right) \\ =\frac{1}{27}-\frac{27}{1} \\ =\frac{1}{27}-\frac{27\times 27}{1\times 27} \\ =\frac{1}{27}-\frac{729}{27} \\ =\frac{1-729}{27} \\ =\frac{-728}{27} \\ \mathrm{Also},b=3^3-3^{-3} \\ =3^3-\frac{1}{3^3}\left(\mathrm{As},x^{-m}=\frac{1}{x^m}\right) \\ =\frac{27}{1}-\frac{1}{27} \\ =\frac{27\times 27}{1\times 27}-\frac{1}{27} \\ =\frac{729}{27}-\frac{1}{27} \\ =\frac{729-1}{27} \\ =\frac{728}{27} \\ \mathrm{Now}, \\ \frac{a}{b}-\frac{b}{a} \\ =\frac{\left({\displaystyle \frac{-728}{27}}\right)}{\left({\displaystyle \frac{728}{27}}\right)}-\frac{\left({\displaystyle \frac{728}{27}}\right)}{\left({\displaystyle \frac{-728}{27}}\right)} \\ =\left(-1\right)-\left(-1\right) \\ =-1+1 \\ =0 \end{gathered}$$

Hence, the correct alternative is option (a).