Áp dụng HĐT \(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\) ta có:
\(a^3=\left(\sqrt[3]{13+7\sqrt{6}}+\sqrt[3]{13-7\sqrt{6}}\right)^3\)
\(=26+3\sqrt[3]{\left(13+7\sqrt{6}\right)\left(13-7\sqrt{6}\right)}.\left(\sqrt[3]{13+7\sqrt{6}}+\sqrt[3]{13-7\sqrt{6}}\right)\)
\(=26-3.\left(-5\right).\left(\sqrt[3]{13+7\sqrt{6}}+\sqrt[3]{13-7\sqrt{6}}\right)\)
\(=26-15.a\)
\(\Rightarrow a^3=26-15a\Leftrightarrow a^3+15a-26=0\)
\(\Leftrightarrow a^3+15a-25=1\)
Vậy \(\left(a^3+15a-25\right)^{2013}=1^{2013}=1\)