d: \(=y^8-2\cdot y^4\cdot\dfrac{1}{3}+\dfrac{1}{9}=\left(y^4-\dfrac{1}{3}\right)^2\)
`1/9 - 2/3 y^4 + y^8`
`=(1/3)^2 - 2.1/3 . y + (y^4)^2`
`=(1/3 - y^4)^2`
\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8\)
\(=\left(\dfrac{1}{3}\right)^2-2\cdot\dfrac{1}{3}.y^4+\left(y^4\right)^2\)
\(=\left(\dfrac{1}{3}-y^4\right)^2\)