Question

Cho :  

         \(x=\frac{1}{3}\left(\sqrt[3]{\frac{23+\sqrt{513}}{4}}+\sqrt[3]{\frac{23-\sqrt{513}}{4}}-1\right)\)

Hãy tính : \(A=x^3+x^2+1\)

Answer 1

Đặt \(a=\sqrt[3]{\frac{23+\sqrt{513}}{4}};b=\sqrt[3]{\frac{23-\sqrt{513}}{4}}\Rightarrow a^3+b^3=\frac{23}{2}\)

\(ab=1\) và \(3x+1=a+b\)

Suy ra : \(\left(3x+1\right)^3-27x^3+27x^2+9+1=27\left(x^3+x^2+1\right)+3\left(3x+1\right)-29\)

hay : \(A=\frac{\left(3x+1\right)^3-3\left(3x+1\right)+29}{27}=\frac{\left(a+b\right)^3-3\left(a+b\right)+29}{27}\)

                                             \(=\frac{a^3+b^3+3ab\left(a+b\right)-3\left(a+b\right)+29}{27}=\frac{\frac{23}{2}+29}{27}=\frac{3}{2}\)

Vậy giá trị của biểu thức đã cho là \(A=\frac{3}{2}\)