Resposta :
Temos que:
[tex](\text{a}+\text{b})^3=(\text{a}+\text{b})^2\cdot(\text{a}+\text{b})[/tex]
Como [tex](\text{a}+\text{b})^2=\text{a}^2+2\text{ab}+\text{b}^2[/tex], temos que:
[tex](\text{a}+\text{b})^3=(\text{a}^2+2\text{ab}+\text{b}^2)\cdot(\text{a}+\text{b})[/tex]
Logo:
[tex](\text{a}+\text{b})^3=\text{a}^3+3\text{a}^2\text{b}+3\text{a}\text{b}^2+\text{b}^3[/tex]
Analogamente, temos:
[tex](\text{a}-\text{b})^3=(\text{a}-\text{b})^2\cdot(\text{a}-\text{b})[/tex]
Como [tex](\text{a}-\text{b})^2=\text{a}^2-2\text{ab}+\text{b}^2[/tex], temos que:
[tex](\text{a}-\text{b})^3=(\text{a}^2-2\text{ab}+\text{b}^2)\cdot(\text{a}-\text{b})[/tex]
Logo:
[tex](\text{a}+\text{b})^3=\text{a}^3-3\text{a}^2\text{b}+3\text{a}\text{b}^2-\text{b}^3[/tex]
(x+a)³=(x+a)(x+a)(x+a)
(x+a)³=(x²+2.a.x+a²)(x+a)
(x+a)³=x³+ax²+2.a.x²+2.a².x+a².x+a³
(x+a)³=x³+3.a.x²+3a².x+a³
(x-a)³=(x-a)(x-a)(x-a)
(x-a)³=(x²-2.a.x+a²)(x-a)
(x-a)³=x³-ax²-2.a.x²+2.a².x+a².x-a³
(x-a)³=x³-3.a.x²+3a².x-a³