Question

Desenvolva as seguintes expressões (a+b)3

Answer 1

• Resposta: $\large\displaystyle\begin{gathered}(a+b)^{3}= a^{3}+3a^{2}b+3ab^{2}+b^{3}\end{gathered}$.

Desejamos desenvolver (a+b)³.

Para desenvolver, irei utilizar o binômio de Newton. Dado por:

                    $\large\displaystyle\begin{gathered} (a+b)^{n}=\sum_{p=0}^{n}\binom{n}{p}a^{n-p}b^{p} \end{gathered}$

• Logo:

$\large\displaystyle\begin{gathered} (a+b)^{n}=\sum_{p=0}^{n}\binom{n}{p}a^{n-p}b^{p} \end{gathered}$

$\large\displaystyle\begin{gathered} (a+b)^{3}=\sum_{p=0}^{3}\binom{3}{0}a^{3-0}b^{0} \end{gathered}$

$\large\displaystyle\begin{gathered} (a+b)^{3}=\binom{3}{0}a^{3-0}b^{0} + \binom{3}{1}a^{3-1}b^{1} +\binom{3}{2}a^{3-2}b^{2}+\binom{3}{3}a^{3-3}b^{3}\end{gathered}$

$\large\displaystyle\begin{gathered} (a+b)^{3}=\binom{3}{0}a^{3}+ \binom{3}{1}a^{2}b+\binom{3}{2}ab^{2}+\binom{3}{3}b^{3}\end{gathered}$

Lembrando que:

                   $\large\displaystyle\begin{gathered} \binom{n}{p}=\frac{n!}{p!(n-p)!} \end{gathered}$

• Ficando então:

$\large\displaystyle\begin{gathered} (a+b)^{3}=\binom{3}{0}a^{3}+ \binom{3}{1}a^{2}b+\binom{3}{2}ab^{2}+\binom{3}{3}b^{3}\end{gathered}$

$\large\displaystyle\begin{gathered} (a+b)^{3}=\frac{3!}{0!(3-0)!} a^{3}+ \frac{3!}{1!(3-1)!}a^{2}b+\frac{3!}{2!(3-2)!}ab^{2}+\frac{3!}{3!(3-3)!}b^{3}\end{gathered}$

$\large\displaystyle\begin{gathered} (a+b)^{3}=\frac{6}{6!} a^{3}+ \frac{6}{2}a^{2}b+\frac{6}{2}ab^{2}+\frac{6}{6}b^{3}\end{gathered}$

$\large\displaystyle\text{ \begin{gathered} \therefore \boxed{\boxed{\green{(a+b)^{3}= a^{3}+3a^{2}b+3ab^{2}+b^{3}}}}\ \checkmark\end{gathered} }$

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