Question
Answer
$$(6+dx)^3-(dx+4)^3=6d^2x^2+60dx+152$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(6+dx)^3-(dx+4)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}216+108dx+18d^2x^2+d^3x^3-(1d^3x^3+12d^2x^2+48dx+64) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}216+108dx+18d^2x^2+d^3x^3-d^3x^3-12d^2x^2-48dx-64 \xlongequal{ } \\[1 em] & \xlongequal{ }216+108dx+18d^2x^2+ \cancel{d^3x^3} -\cancel{d^3x^3}-12d^2x^2-48dx-64 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}6d^2x^2+60dx+152\end{aligned} $$ | |
| ① | Find $ \left(6+dx\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 6 $ and $ B = dx $. $$ \left(6+dx\right)^3 = 6^3+3 \cdot 6^2 \cdot dx + 3 \cdot 6 \cdot \left( dx \right)^2+\left( dx \right)^3 = 216+108dx+18d^2x^2+d^3x^3 $$Find $ \left(dx+4\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = dx $ and $ B = 4 $. $$ \left(dx+4\right)^3 = \left( dx \right)^3+3 \cdot \left( dx \right)^2 \cdot 4 + 3 \cdot dx \cdot 4^2+4^3 = d^3x^3+12d^2x^2+48dx+64 $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( d^3x^3+12d^2x^2+48dx+64 \right) = -d^3x^3-12d^2x^2-48dx-64 $$ |
| ③ | Combine like terms: $$ \color{blue}{216} + \color{red}{108dx} + \color{green}{18d^2x^2} + \, \color{orange}{ \cancel{d^3x^3}} \, \, \color{orange}{ -\cancel{d^3x^3}} \, \color{green}{-12d^2x^2} \color{red}{-48dx} \color{blue}{-64} = \color{green}{6d^2x^2} + \color{red}{60dx} + \color{blue}{152} $$ |
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