Đặt \(x=a+b=\sqrt[3]{7+\sqrt[]{50}}+\sqrt[3]{7-\sqrt[]{50}}\)
\(\Rightarrow x^3=14+3\left(\sqrt[3]{7+\sqrt[]{50}}+\sqrt[3]{7-\sqrt[]{50}}\right)\sqrt[3]{\left(7+\sqrt[]{50}\right)\left(7-\sqrt[]{50}\right)}\)
\(\Rightarrow x^3=14+3x.\left(-1\right)\)
\(\Rightarrow x^3+3x-14=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)=0\)
\(\Rightarrow x-2=0\) hay \(a+b=2\)
//Lại có: \(ab=\sqrt[3]{\left(7+\sqrt[]{50}\right)\left(7-\sqrt[]{50}\right)}=-1\)
\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=6\)
\(\Rightarrow a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=34\)
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=14\)
\(\Rightarrow a^7+b^7=\left(a^3+b^3\right)\left(a^4+b^4\right)-\left(ab\right)^3\left(a+b\right)=478\)
