Lời giải:
Đặt \(\sqrt[3]{26+15\sqrt{3}}=a; \sqrt[3]{26-15\sqrt{3}}=b\)
\(\Rightarrow \left\{\begin{matrix} a^3+b^3=52\\ ab=\sqrt[3]{(26+15\sqrt{3})(26-15\sqrt{3})}=\sqrt[3]{1}=1\end{matrix}\right.\)
Khi đó:
\((\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}})^3=(a+b)^3=a^3+b^3+3ab(a+b)\)
\(\Leftrightarrow (a+b)^3=52+3(a+b)\). Đặt \(a+b=t\)
\(\Leftrightarrow t^3-3t-52=0\)
\(\Leftrightarrow t(t^2-16)+13(t-4)=0\)
\(\Leftrightarrow (t-4)(t^2+4t+13)=0\)
Thấy rằng \(t^2+4t+13=(t+2)^2+9>0\forall t\) nên \(t-4=0\Leftrightarrow t=4\)
Vậy giá trị của biểu thức là $4$
\(a=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)
\(a^3=26+15\sqrt{3}+26-15\sqrt{3}+3\sqrt[3]{\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)}a\)
\(a^3=52+3.\sqrt{26^2-3.15^2}.a\)
\(a^3=52+3a\)
\(a^3-64=3a-12\)
\(a^3-4^3=3\left(a-4\right)\)
\(\left(a-4\right)\left(a^2+4a+16\right)=3\left(a-4\right)\)
\(\left(a-4\right)\left(a^2+4a+13\right)=0\) \(\Delta_{a^2+4a+13}=4-13< 0=>vn\)
a=4 là duy nhất
Ta có \(\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{8+3.4.\sqrt{3}+3.2.3+3\sqrt{3}}+\sqrt[3]{8-3.4.\sqrt{3}+3.2.3-3\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{\left(2-\sqrt{3}\right)^3}=2+\sqrt{3}+2-\sqrt{3}=4\)
