Lời giải của Nguyễn Thị Ngọc Thơ

Đặt \(x=\sqrt[3]{8+3\sqrt[]{21}}+\sqrt[3]{8-3\sqrt[]{21}}\)

\(\Rightarrow x^3=8+3\sqrt[]{21}+8-3\sqrt[]{21}+3\sqrt[3]{\left(8+3\sqrt[]{21}\right)\left(8-3\sqrt[]{21}\right)}\left(\sqrt[3]{8+3\sqrt[]{21}}+\sqrt[3]{8-3\sqrt[]{21}}\right)\)

\(\Rightarrow x^3=16+3.\sqrt[3]{-125}.x\)

\(\Rightarrow x^3=16-15x\)

\(\Rightarrow x^3+15x-16=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+x+16\right)=0\)

\(\Rightarrow x=1\) (do \(x^2+x+16=\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}>0;\forall x\))

Vậy \(\sqrt[3]{8+3\sqrt[]{21}}+\sqrt[3]{8-3\sqrt[]{21}}=1\)