Question

Factorise: $a^2+\frac{1}{a^2}+3-2a-\frac{2}{a}$

• A. $(a+\frac{1}{a}-1)(a-\frac{1}{a}+1)$
• B. $(a+\frac{1}{a}-1)(a+\frac{1}{a}+1)$
• C. $(a+\frac{1}{a}+1)(a+\frac{1}{a}+1)$
• D. $(a+\frac{1}{a}-1)(a+\frac{1}{a}-1)$

Answer 1

The correct option is D $(a+\frac{1}{a}-1)(a+\frac{1}{a}-1)$

$a^2+\frac{1}{a^2}+3-2a-\frac{2}{a}$
$=a^2+\frac{1}{a^2}+1+2-2a-\frac{2}{a}$
$={\left(a\right)}^2+{\left(\frac{1}{a}\right)}^2+{(-1)}^2+2\times a\times \frac{1}{a}+2\times a\times (-1)+2\times \frac{1}{a}\times (-1)$
$\because {(a+b-c)}^2=a^2+b^2+c^2+2ab-2ac-2cb$
$={(a+\frac{1}{a}-1)}^2=(a+\frac{1}{a}-1)(a+\frac{1}{a}-1)$