\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
Với \(a=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}};\text{ }b=\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\) \(\left(x^2\ge4\right)\)
Thì \(a^3+b^3=x^3-3x\)'
\(a.b=\sqrt[3]{\frac{\left(x^3-3x\right)^2-\left(x^2-1\right)^2\left(x^2-4\right)}{4}}=\sqrt[3]{\frac{4}{4}}=1\)
Suy ra: \(B^3=x^3-3x+3.\left(a+b\right)\)
+Xét trường hợp \(a+b=0\Leftrightarrow\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}=-\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)
\(\Leftrightarrow\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}=-\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}\)
\(\Leftrightarrow x^3-3x=0\Leftrightarrow x\left(x^2-3\right)=0\text{ (vô nghiệm }\text{do }x^2\ge4\text{)}\)
+Vậy \(a+b\ne0\text{ }\forall x^2\ge4\), thay \(a+b=B\), ta được:
\(B^3=x^3-3x+3B\)
\(\Leftrightarrow B^3-x^3-3\left(B-x\right)=0\)
\(\Leftrightarrow\left(B-x\right)\left(B^2+x^2+Bx-3\right)=0\)
\(\Leftrightarrow B=x\text{ hoặc }\left(B+\frac{x}{2}\right)^2+\frac{3x^2}{4}-3=0\text{ (1)}\)
Mà \(x^2\ge4\text{ nên }\frac{3x^2}{4}\ge3\Rightarrow\left(B+\frac{x}{2}\right)^2+\frac{3x^2}{4}-3\ge0\)
Lại có: \(x=2\text{ thì }B=\sqrt[3]{\frac{2^3-3.2}{2}}+\sqrt[3]{\frac{2^3-3.2}{2}}=2\)
\(x=-2\text{ thì }B=\sqrt[3]{\frac{\left(-2\right)^3-3.\left(-2\right)}{2}}+\sqrt[3]{\frac{\left(-2\right)^3-3.\left(-2\right)}{2}}=-2\)
Nên \(B+\frac{x}{2}=0\text{ và }x^2=4\text{ không đồng thời xảy ra, hay }\left(B+\frac{x}{2}\right)^2+\frac{3x^2}{4}-3>0\)
Vậy (1) vô nghiệm.
Do đó \(B=x\)
Vậy \(B=x\text{ }\forall x^2\ge4\)
