Question
Factorize the following expressions:
$\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\text{a}+\frac{2}{\text{a}}$
$\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\text{a}+\frac{2}{\text{a}}$
✓
Answer
$=\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}^3-\Big(\frac{1}{\text{a}^3}\Big)\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\Big(\frac{1}{\text{a}}\Big)^2\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$\left[\therefore \mathrm{a}^3-\mathrm{b}^3=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2\right)\right]$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
$\therefore\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+2\text{a}$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
$=\Big(\text{a}^3-\Big(\frac{1}{\text{a}^3}\Big)\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\Big(\frac{1}{\text{a}}\Big)^2\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$\left[\therefore \mathrm{a}^3-\mathrm{b}^3=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2\right)\right]$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
$\therefore\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+2\text{a}$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
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