Answer
$$(x-y)^3=x^3-3x^2y+3xy^2-y^3$$
Explanation
| $$ \begin{aligned}(x-y)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^3-3x^2y+3xy^2-y^3\end{aligned} $$ | |
| 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 $. $$ \left(x-y\right)^3 = x^3-3 \cdot x^2 \cdot y + 3 \cdot x \cdot y^2-y^3 = x^3-3x^2y+3xy^2-y^3 $$ |
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