Question

Factorise the following: $a^3+4-\frac{1}{a^3}$

• A. $(a-1-\frac{1}{a})(a^2+1+\frac{1}{a^2}-a+\frac{1}{a}+1)$
• B. $(a+1-\frac{1}{a})(a^2+1+\frac{1}{a^2}-a+\frac{1}{a}-1)$
• C. $(a-1-\frac{1}{a})(a^2+1+\frac{1}{a}-a+\frac{1}{a}-1)$
• D. $(a+1-\frac{1}{a})(a^2+1+\frac{1}{a^2}-a+\frac{1}{a}+1)$

Answer 1

The correct option is D $(a+1-\frac{1}{a})(a^2+1+\frac{1}{a^2}-a+\frac{1}{a}+1)$

$a^3+4-\frac{1}{a^3}$

$=a^3+1-\frac{1}{a^3}+3$

$=a^3+1^3-(\frac{1}{a}{)}^3-3(a\left)\right(1\left)\right(\frac{1}{a})$

Using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we get,

$=(a+1-\frac{1}{a})(a^2+1^2+(\frac{1}{a}{)}^2-a\left(1\right)-1(-\frac{1}{a})-a(-\frac{1}{a}))$

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