Question : $3\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]+\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]^3=?$
Option 1: $a^2-\frac{1}{a^3}$
Option 2: $a^3-\frac{1}{a^3}$
Option 3: $a^3+\frac{1}{a^3}$
Option 4: $a^2-\frac{1}{a^2}$
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Correct Answer: $a^3-\frac{1}{a^3}$
Solution : $3\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]+\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]^3$ Using $(a-b)^3=a^3 - b^3 -3ab(a-b)$ $=3\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right] + \left[\mathrm{a^3}-\frac{1}{\mathrm{a^3}}-3\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]\right]$ $=3\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right] + \left[\mathrm{a^3}-\frac{1}{\mathrm{a^3}}\right]-3\left[\mathrm{a}-\frac{1}{\mathrm{a}}\right]$ $=\left[\mathrm{a^3}-\frac{1}{\mathrm{a^3}}\right]$ Hence, the correct answer is $\mathrm{a^3}-\frac{1}{\mathrm{a^3}}$.
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