Question

Tính giá trị của biểu thức : \(Q=x^3+ax+b\)  biết \(x=\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}\)\(+\sqrt[3]{\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}\)

Answer 1

Ta có: \(x=\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}+\sqrt[3]{\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}\)

=> \(x^3=\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+3\cdot\sqrt[3]{\left(\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)\left(\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)}\cdot x.\)

    =  \(-b+\sqrt[3]{\frac{b^2}{4}-\left(\frac{b^2}{4}+\frac{a^3}{27}\right)}\cdot x\)

=\(-b+\sqrt[3]{\frac{a^3}{27}}\cdot x=-b+\frac{a}{27}\cdot x\)

=> \(x^3+b=\frac{a}{27}\cdot x\)

Vậy  \(x^3+ax+b=\frac{a}{27}\cdot x+ax=\frac{28a}{27}\cdot x\)