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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