Question

Докажите тождество

(b+1)^3-3(b+1)^2+3(b+1)-1=b^3

a^4+5a^3+9a^2+7a+2=(a+1)^3(a+2)

Answer 1

$(b+1)^3-3(b+1)^2+3(b+1)-1=b^3$

$b^3+3b^2+3b+1-3(b^2+2b+1)+3(b+1)-1=b^3$ 

$b^3+3b^2+3b+1-3b^2-6b-3+3b+3-1=b^3$ 

$b^3+3b^2+3b+1-3b^2-6b-3+3b+3-1=b^3$ 

$b^3=b^3$

 

 $a^4+5a^3+9a^2+7a+2=(a+1)^3(a+2)$

$a^4+5a^3+9a^2+7a+2=(a^3+3a^2+3a+1)(a+2)$

$a^4+5a^3+9a^2+7a+2=$ 

$=a^4+3a^3+3a^2+a+2a^3+6a^2+6a+2$ 

 $a^4+5a^3+9a^2+7a+2=a^4+5a^3+9a^2+7a+2$