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