(a+b)^3 ?
Let a = x:
If (x+b)3=0 then (x+b)(x+b)(x+b)=0 is a Polynom with the roots x1=−bx2=−bx3=−b.
Vieta's formulas:
If x3+Ax2+Bx+C=0, then
A=−(x1+x2+x3)B=x1x2+x1x3+x2x3C=−(x1x2x3)A=−[(−b)+(−b)+(−b)]=−[3(−b)]=3bB=(−b)(−b)+(−b)(−b)+(−b)(−b)=3b2C=−[(−b)(−b)(−b)]=−[−3b3]=b3
so we have:
(x+b)3=x3+Ax2+Bx+C=0(x+b)3=x3+(3b)x2+(3b2)x+(b3)|x=a(a+b)3=a3+(3b)a2+(3b2)a+(b3)(a+b)3=a3+3a2b+3ab2+b3
(a+b)^3 ?
Let a = x:
If (x+b)3=0 then (x+b)(x+b)(x+b)=0 is a Polynom with the roots x1=−bx2=−bx3=−b.
Vieta's formulas:
If x3+Ax2+Bx+C=0, then
A=−(x1+x2+x3)B=x1x2+x1x3+x2x3C=−(x1x2x3)A=−[(−b)+(−b)+(−b)]=−[3(−b)]=3bB=(−b)(−b)+(−b)(−b)+(−b)(−b)=3b2C=−[(−b)(−b)(−b)]=−[−3b3]=b3
so we have:
(x+b)3=x3+Ax2+Bx+C=0(x+b)3=x3+(3b)x2+(3b2)x+(b3)|x=a(a+b)3=a3+(3b)a2+(3b2)a+(b3)(a+b)3=a3+3a2b+3ab2+b3