Absolute value polynomial

Since absolute value is positive even if what is inside the absolute value brackets is negative, and $2y-x=-(x-2y)$ , we can convert all the $(2y-x)$ into $(x-2y)$ . Then, group the 1st, 2nd, and 4th term to form a quadratic and the 3rd and 5th term to form a difference of squares . Let $a=|x-2y|^3$ to make the math easier. The quadratic can be factored into $(7a-109)(a+27)$ and the difference of squares into $(a+45)(a+27)$ . An $(a+27)$ can be factored out, leaving $(a+27)(8a-64)$ . Since $a$ represents a cube, $|x-2y|^3$ can be substituted back into the equation. Next, the Cube Sum and Difference formulas can be applied to get a product of two binomials and two trinomials. The Zero-Product Property can be applied here. Since absolute value cannot be negative, this eliminates one of the binomials. Also, the two trinomials have no solutions, which leaves us with $(|x-2y|)=2$ . This can be split into $x-2y=2$ and $x-2y=-2$ , and by adding $2y$ , subtracting $2$ for the 1st equation and $-2$ for the second equation, and dividing both sides by $2y$ , we can obtain that $a=2$ . As a result, $a=2$ is the final answer.