\(x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Rightarrow x^3=5+2\sqrt{13}+5-2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}.x\)
\(=10+3x\sqrt[3]{25-52}\)
\(=10+3x\sqrt[3]{-27}\)
\(=10-9x\)
\(\Rightarrow x^3+9x-10=0\)
\(\Leftrightarrow x^3-x+10x-10=0\)
\(\Leftrightarrow x\left(x^2-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+10\right)=0\)
Vì \(x^2+x+10=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}>0\forall x\)
=> x - 1 = 0
=> x = 1
Thay vào A = 12015 - 12016 = 0
Vậy A = 0
