\(A=\sqrt[3]{5-2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\)
Đặt \(a=\sqrt[3]{5-2\sqrt{13}},b=\sqrt[3]{5+2\sqrt{13}}\).
Suy ra \(a^3+b^3=10\) và \(a.b=\sqrt[3]{5-2\sqrt{13}}.\sqrt[3]{5+2\sqrt{13}}=\sqrt[3]{-27}=-3\).
Ta có
\(A^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)\(=10+3.\left(-3\right)\left(a+b\right)\).
Suy ra:
\(\left(a+b\right)^3+9\left(a+b\right)-10=0\)\(\Leftrightarrow\left(a+b\right)^3-\left(a+b\right)+10\left(a+b\right)-10=0\)
\(\Leftrightarrow\left(a+b\right).\left[\left(a+b\right)^2-1\right]+10\left(a+b-1\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+b-1\right)\left(a+b+1\right)+10\left(a+b-1\right)=0\)
\(\Leftrightarrow\left(a+b-1\right)\left[\left(a+b\right)\left(a+b+1\right)+10\right]=0\)
\(\Leftrightarrow\left(a+b-1\right)\left[\left(a+b\right)^2+\left(a+b\right)+10\right]=0\)
\(\Leftrightarrow a+b-1=0\) (do \(\left(a+b\right)^2+\left(a+b\right)+10>0,\forall a+b\).
\(\Leftrightarrow a+b=1\)
Hay \(A=\sqrt[3]{5-2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}=1\).
