3(b-1)² +(b+2)(b²-2b+4)-(b+1)³ при b=⅓​

Ответ:

7

Объяснение:

3(b-1)² +(b+2)(b²-2b+4)-(b+1)³ = 3b² - 6b + 3 + b³ + 2b² - 2b² - 4b + 4b + 8 - b³ - 3b² - 3b - 1 =

= -9b + 10 = -9*1/3 + 10 = -3 + 10 = 7

3(b-1)² +(b+2)(b²-2b+4)-(b+1)³=3(b-1)² +(b+2)((b-1)²+3)-(b+1)³=

(b-1)²(3 +b+2)+(3b+6)-(b+1)³=(b-1)²(5 +b)+(3b+6) -(b+1)³

При b=1/3, ((-2/3)²)(5+1/3)+(7)-(4/3)³=(4/9)*(16/3)+7-64/27=

(64/27)+7-64/27=7