Question

Qual o volume de um cubo que a aresta mede (2x +1)?

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Volume~do~cubo~de~aresta~\ell}\\\sf V=\ell^3\end{array}}$

$\Large\boxed{\begin{array}{l}\sf \ell=2x+1\\\sf V=\ell^3\\\sf V=(2x+1)^3\\\sf V=(2x)^3+3\cdot(2x)^2\cdot1+3\cdot2x\cdot 1^2+1^3\\\sf V= 8x^3+3\cdot4x^2+6x+1\\\sf V=8x^3+12x^2+6x+1\end{array}}$

Answer 2

✅ Após ter realizado todos os cálculos concluímos que o volume do cubo é:

   $\large\displaystyle\begin{gathered}V = 8x^{3} + 12x^{2} + 6x + 1 \end{gathered}$

O volume do cubo é o cubo da medida da aresta, ou seja:

                  $\large\displaystyle\begin{gathered}V = a^{3} \end{gathered}$

Se a medida da aresta é:

                 $\large\displaystyle\begin{gathered}a = 2x + 1 \end{gathered}$

Então:

        $\large\displaystyle\begin{gathered}V = (2x + 1)^{3} \end{gathered}$

            $\large\displaystyle\begin{gathered}= (2x + 1)\cdot(2x + 1)\cdot(2x + 1) \end{gathered}$

           $\large\displaystyle\begin{gathered}= (4x^{2} + 4x + 1)\cdot(2x + 1) \end{gathered}$

           $\large\displaystyle\begin{gathered}= 8x^{3} + 12x^{2} + 6x + 1 \end{gathered}$

✅ Portanto, o volume do cubo é:

      $\large\displaystyle\begin{gathered}V = 8x^{3} + 12x^{2} + 6x + 1 \end{gathered}$

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