Tap the blue circles to see an explanation.
$$ \begin{aligned}x((1-x)^3-(-x)^3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x(1-3x+3x^2-x^3--x^3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x(3x^2-3x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}3x^3-3x^2+x\end{aligned} $$ | |
① | Find $ \left(1-x\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = 1 $ and $ B = x $. $$ \left(1-x\right)^3 = 1^3-3 \cdot 1^2 \cdot x + 3 \cdot 1 \cdot x^2-x^3 = 1-3x+3x^2-x^3 $$$$ \left( -x \right)^3 = (-1)^3x^3 = -x^3 $$ |
② | Combine like terms: $$ 1-3x+3x^2 \, \color{blue}{ -\cancel{x^3}} \,+ \, \color{blue}{ \cancel{x^3}} \, = 3x^2-3x+1 $$ |
③ | Multiply $ \color{blue}{x} $ by $ \left( 3x^2-3x+1\right) $ $$ \color{blue}{x} \cdot \left( 3x^2-3x+1\right) = 3x^3-3x^2+x $$ |