Answer:
Question:
[tex]\textsf{factor : \ } x^4+\dfrac{1}{x^2}[/tex]
Rewrite the expression using the exponent rule [tex]\dfrac{a^b}{a^c}=a^{(b-c)}[/tex]
[tex]\implies x^4+\dfrac{1}{x^2}=\dfrac{x^6}{x^2}+\dfrac{1}{x^2}[/tex]
Factor out common term [tex]\dfrac{1}{x^2}[/tex] :
[tex]\implies \dfrac{1}{x^2}(x^6+1)}[/tex]
Factor [tex](x^6+1)[/tex]:
Apply the exponent rule [tex](a^b)^c=a^{bc}[/tex] to get cubic exponents:
[tex]\implies x^6+1 = (x^2)^3+1^3[/tex]
Apply the sum of cubes formula: [tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]
where [tex]a=x^2[/tex] and [tex]b=1[/tex]:
[tex]\implies (x^2)^3+1^3=(x^2+1)((x^2)^2-x^2 \cdot 1+1^2)[/tex]
[tex]\implies (x^2+1)(x^4-x^2+1)[/tex]
Therefore,
[tex]\implies \dfrac{1}{x^2}(x^6+1)}=\dfrac{1}{x^2}(x^2+1)(x^4-x^2+1)[/tex]
Simplify by dividing the 2nd parentheses by [tex]x^2[/tex]:
[tex]\implies (x^2+1)\left(\dfrac{x^4-x^2+1}{x^2}\right)[/tex]
[tex]\implies (x^2+1)\left(x^2-1+\dfrac{1}{x^2}\right)[/tex]
Factor out [tex]x[/tex] from first parentheses:
[tex]\implies x\left(x+\dfrac{1}{x}\right)\left(x^2-1+\dfrac{1}{x^2}\right)[/tex]
