Question

Which of the following can be the factor(s) of $(a^2+\frac{1}{a^2}-2-3a+\frac{3}{a})$ ?

• A. $(\frac{1}{a}-a)\left[\right(\frac{1}{a}-a)+3]$
• B. $(a+\frac{1}{a})\left[\right(a+\frac{1}{a})-3]$
• C. $(a-\frac{1}{a})\left[\right(a-\frac{1}{a})-3]$
• D. $(a+\frac{1}{a})\left[\right(a-\frac{1}{a})+3]$

Answer 1

Answer: (A), (C)

The correct options are
A $(\frac{1}{a}-a)\left[\right(\frac{1}{a}-a)+3]$
C $(a-\frac{1}{a})\left[\right(a-\frac{1}{a})-3]$

$\Rightarrow (a^2+\frac{1}{a^2}-2-3a+\frac{3}{a})$

Group the terms as shown,

$\Rightarrow (a^2+\frac{1}{a^2}-2)-(3a-\frac{3}{a})$

$\Rightarrow (a-\frac{1}{a}{)}^2-3(a-\frac{1}{a})$

$\Rightarrow (a-\frac{1}{a})\left[\right(a-\frac{1}{a})-3]$

Taking (-1) common from both the terms, we get,

$\Rightarrow (-1)(-a+\frac{1}{a})(-1)\left[\right(-a+\frac{1}{a})+3]$

$\Rightarrow (\frac{1}{a}-a)\left[\right(\frac{1}{a}-a)+3]$