Question

6(y+1)^2+2(y-1)(y^2+y+1)-2(y+1)^3=32

Answer 1

Ответ:

5

Объяснение:

6(y+1)^2+2(y-1)(y^2+y+1)-2(y+1)^3=32

6(y+1)²+2(y³-1)-2(y+1)³=32

6(y²+2y+1)+2y³-2-2(y³+3y²+3y+1)=32

6y²+12y+6+2y³-2-2y³-6y²-6y-2=32

6y+2=32

6y+2-32=0

6y-30=0

6y=30

y=5