\(\dfrac{1}{a}=\dfrac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\dfrac{\sqrt[3]{2}-1}{\left(\sqrt[3]{2}-1\right)\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}=\dfrac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1\)
\(\Rightarrow\dfrac{1}{a}+1=\sqrt[3]{2}\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{3}{a^2}+\dfrac{3}{a}=\left(\dfrac{1}{a}+1\right)^3-1=2-1=1\)
`a = ` $\sqrt[3]{4}$ `+` $\sqrt[3]{2}$ `+` `1`
`<=> a - 1 = ` $\sqrt[3]{4}$ `+` $\sqrt[3]{2}$ `
`<=> ( a - 1)^3 = `(` $\sqrt[3]{4}$ `+` $\sqrt[3]{2}$ `)^3`
`<=> ( a - 1 ) ^ 3 = 4 + 2 + 3` $\sqrt[3]{8}$ `. ( a - 1 )`
`<=> ( a - 1 ) ^ 3 = 6 + 6 ( a - 1)`
`<=> a^3 - 3a^2 + 3a - 1 - 6 - 6(a-1) = 0`
`<=> a^3 - 3a^2 - 3a - 1= 0`
`<=> a^3 = 3a^2 + 3a + 1`
Ta có:
`3/a + 3/(a^2) + 1/(a^3)`
`= (a^3)/(a^3)`
`= 1`
`color{black}{#Fensiyagi}`