Question
Answer
$$(x+y)^3\cdot2-(x-1)^2z+3x^2+4y=2x^3+6x^2y-x^2z+6xy^2+2y^3+3x^2+2xz+4y-z$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+y)^3\cdot2-(x-1)^2z+3x^2+4y& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^3+3x^2y+3xy^2+y^3)\cdot2-(x^2-2x+1)z+3x^2+4y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2x^3+6x^2y+6xy^2+2y^3-(x^2z-2xz+z)+3x^2+4y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2x^3+6x^2y+6xy^2+2y^3-x^2z+2xz-z+3x^2+4y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2x^3+6x^2y-x^2z+6xy^2+2y^3+3x^2+2xz+4y-z\end{aligned} $$ | |
| ① | Find $ \left(x+y\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = x $ and $ B = y $. $$ \left(x+y\right)^3 = x^3+3 \cdot x^2 \cdot y + 3 \cdot x \cdot y^2+y^3 = x^3+3x^2y+3xy^2+y^3 $$Find $ \left(x-1\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(x-1\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 1 + \color{red}{1^2} = x^2-2x+1\end{aligned} $$ |
| ② | $$ \left( \color{blue}{x^3+3x^2y+3xy^2+y^3}\right) \cdot 2 = 2x^3+6x^2y+6xy^2+2y^3 $$$$ \left( \color{blue}{x^2-2x+1}\right) \cdot z = x^2z-2xz+z $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( x^2z-2xz+z \right) = -x^2z+2xz-z $$ |
| ④ | Combine like terms: $$ 2x^3+6x^2y+6xy^2+2y^3-x^2z+2xz-z+3x^2+4y = 2x^3+6x^2y-x^2z+6xy^2+2y^3+3x^2+2xz+4y-z $$ |
This page was created using
Simplify polynomials calculator