Question

(2у-1)в кубе+(2у+1)в кубе=у(1+16у в квадрате)+2 решить

Answer 1

(a-b)³=a³-3a²b+3ab²-b³      (a+b)³=a³+3a²b+3ab²+b³

(2y-1)³+(2y+1)³=y(1+16y²)+2
((2y)³-3*(2y)²*1+3*2y*1²-1³)+((2y)³+3*(2y)²*1+3*2y*1²+1³)=y+16y³+2
8y³-12y²+6y-1+8y³+12y²+6y+1=y+16y³+2
16y³+12y=y+16y³+2
16y³+12y-y-16y³=2
11y=2
y=2/11