B2. In the set of real numbers, solve the equation $\frac{x^{6}-5 x^{3}}{14}=1$ exactly.
B2. We rearrange the equation: $x^{6}-5 x^{3}-14=0$. We introduce a new variable, for example, $y=x^{3}$ and write the new equation: $y^{2}-5 y-14=0$. We solve the quadratic equation, for example, by factoring using Vieta's formulas: $(y-7)(y+2)=0$. The solutions to the equation with the new variable are $y_{1}=7$ and $y_{2}=-2$. We substitute the new variable: $x^{3}=7$ and $x^{3}=-2$. We write the exact solutions to the equation: $x_{1}=\sqrt[3]{7}$ and $x_{2}=-\sqrt[3]{2}$.

Rearrangement of the equation $x^{6}-5 x^{3}-14=0$.....................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_37ea02f5a530d200baf4g-24.jpg?height=57&width=1736&top_left_y=2690&top_left_x=177)
Writing the equation with the new variable $y^{2}-5 y-14=0$ ..... 1 point
Solving the equation, for example, using Vieta's formulas $(y-7)(y+2)=0$ ..... $1^{*}$ point
Writing the solutions $y_{1}=7$ and $y_{2}=-2$ ..... 1 point
Substitution of the new variables: $x^{3}=7$ and $x^{3}=-2$ ..... 1 point
Writing the exact solutions to the equation: $x_{1}=\sqrt[3]{7}$ and $x_{2}=-\sqrt[3]{2}$. ..... $1+1$ points