Решите уравнение1)(2у-1)^3+(2y+1)^3=y(1+16y^2)+2

(2y-1)³+(2y+1)³=y(1+16y²)+2
((2y-1)+(2y+1))((2y-1)²-(2y-1)(2y+1)+(2y+1)²)=y+16y³+2
(2y-1+2y+1)(4y²-4y+1-(4y²-1)+(4y²+4y+1))=y+16y³+2
4y(4y²-4y+1-4y²+1+4y²+4y+1)=y+16y³+2
4y(4y²+3)=y+16y³+2
16y³+12y=y+16y³+2
16y³+12y-y-16y³=2
11y=2
y=2/11