(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 $$x_1 = -b\qquad x_2 = -b\qquad x_3 = -b$$.
Vieta's formulas:
If $$x^3+Ax^2+Bx+C=0$$, then
$$\\\small{\text{$
A =-(x_1 +x_2 +x_3 )\qquad
B =x_1 x_2 +x_1 x_3 +x_2 x_3 \qquad
C =-(x_1 x_2 x_3 )
$}}\\\\
\small{\text{$
\begin{array}{rcl}
A &=& -[ (-b) +(-b) +(-b) ] = -[3(-b)]=3b\\
B &=& (-b) (-b) +(-b) (-b) +(-b) (-b)=3b^2 \\
C &=& -[(-b) (-b) (-b) ]=-[-3b^3]=b^3
\end{array}
$}}\\$$
so we have:
$$\\ (x+b)^3= x^3+Ax^2+Bx+C=0 \\
(x+b)^3= x^3+(3b)x^2+(3b^2)x+(b^3) \qquad | \qquad \textcolor[rgb]{1,0,0}{x = a}\\
(a+b)^3= a^3+(3b)a^2+(3b^2)a+(b^3)\\\\
\mathbf{(a+b)^3= a^3+3a^2b+3ab^2+b^3}$$
(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 $$x_1 = -b\qquad x_2 = -b\qquad x_3 = -b$$.
Vieta's formulas:
If $$x^3+Ax^2+Bx+C=0$$, then
$$\\\small{\text{$
A =-(x_1 +x_2 +x_3 )\qquad
B =x_1 x_2 +x_1 x_3 +x_2 x_3 \qquad
C =-(x_1 x_2 x_3 )
$}}\\\\
\small{\text{$
\begin{array}{rcl}
A &=& -[ (-b) +(-b) +(-b) ] = -[3(-b)]=3b\\
B &=& (-b) (-b) +(-b) (-b) +(-b) (-b)=3b^2 \\
C &=& -[(-b) (-b) (-b) ]=-[-3b^3]=b^3
\end{array}
$}}\\$$
so we have:
$$\\ (x+b)^3= x^3+Ax^2+Bx+C=0 \\
(x+b)^3= x^3+(3b)x^2+(3b^2)x+(b^3) \qquad | \qquad \textcolor[rgb]{1,0,0}{x = a}\\
(a+b)^3= a^3+(3b)a^2+(3b^2)a+(b^3)\\\\
\mathbf{(a+b)^3= a^3+3a^2b+3ab^2+b^3}$$