Answer:
[tex]\large\boxed{\left(x^4+\dfrac{1}{9}\right)\left(x^2+\dfrac{1}{3}\right)\left(x+\dfrac{\sqrt3}{3}\right)\left(x-\dfrac{\sqrt3}{3}\right)}[/tex]
Step-by-step explanation:
[tex]Use\\\\a^2-b^2=(a-b)(a+b)\\\\(a^n)^m=a^{nm}\\\\x^8-\dfrac{1}{81}=x^{4\cdot2}-\dfrac{1}{9^2}=(x^4)^2-\left(\dfrac{1}{9}\right)^2=\left(x^4+\dfrac{1}{9}\right)\underbrace{\left(x^4-\dfrac{1}{9}\right)}_{(*)}\\\\(*)=x^{2\cdot2}-\dfrac{1}{3^2}=(x^2)^2-\left(\dfrac{1}{3}\right)^2=\left(x^2+\dfrac{1}{3}\right)\underbrace{\left(x^2-\dfrac{1}{3}\right)}_{(**)}\\\\(**)=x^2-\dfrac{3}{9}=x^2-\dfrac{(\sqrt3)^2}{3^2}=x^2-\left(\dfrac{\sqrt3}{3}\right)^2=\left(x+\dfrac{\sqrt3}{3}\right)\left(x-\dfrac{\sqrt3}{3}\right)[/tex]
