Đáp án: $ P=2011^{2016}$
Giải thích các bước giải:
Ta có:
$a=\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}}$
$\to a^3=(\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}})^3$
$\to a^3=38+17\sqrt{5}+38-17\sqrt{5}+3\cdot \sqrt[3]{38+17\sqrt{5}}\cdot \sqrt[3]{38-17\sqrt{5}}( \sqrt[3]{38+17\sqrt{5}}+ \sqrt[3]{38-17\sqrt{5}})$
$\to a^3=76+3\cdot \sqrt[3]{(38+17\sqrt{5})(38-17\sqrt{5})}( \sqrt[3]{38+17\sqrt{5}}+ \sqrt[3]{38-17\sqrt{5}})$
$\to a^3=76+3\cdot \sqrt[3]{38^2-(17\sqrt{5})^2}( \sqrt[3]{38+17\sqrt{5}}+ \sqrt[3]{38-17\sqrt{5}})$
$\to a^3=76+3\cdot \sqrt[3]{-1}\cdot a$
$\to a^3=76-3a$
$\to a^3+3a=76$
$\to P=(a^3+3a+1935)^{2016}$
$\to P=(76+1935)^{2016}$
$\to P=2011^{2016}$
