Solve the following equations: $3x^{3} - 8xy^{2} + y^{3} + 21 = 0, x^{2}(y - x) = 1$
- $1, 2$
- $-1, 3$
- $, \sqrt [3]{\dfrac {1}{2}}, 3\sqrt [3]{\dfrac {1}{2}}$.
- $\sqrt {\dfrac {3}{2}}, \dfrac {1}{2}$
From the second equation x squared multiplied by (y minus x) equals 1, we can factor it as x(xy minus x squared) equals 1. Let y equal kx, which makes the equation x cubed multiplied by (k minus 1) equal 1, so x cubed equals 1 divided by (k minus 1). Multiplying the entire second original equation by 3y gives 3x squared y squared minus 3x cubed y equals 3y. Subtracting this from the first equation 3x cubed minus 8xy squared plus y cubed plus 21 equals 0 eliminates the x cubed terms, leaving 3x cubed minus 11xy squared plus y cubed plus 21 minus 3y equals 0. Substituting y equals kx into this relation and solving the system of equations yields the ratio of y to x as 3. Substituting y equals 3x back into the second equation gives x squared(3x minus x) equals 1, which simplifies to 2x cubed equals 1, meaning x equals the cube root of one half and y equals 3 times the cube root of one half.