Question
Answer
$$(y^2-x^2)^3=-x^6+3x^4y^2-3x^2y^4+y^6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(y^2-x^2)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}y^6-3x^2y^4+3x^4y^2-x^6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-x^6+3x^4y^2-3x^2y^4+y^6\end{aligned} $$ | |
| ① | Find $ \left(y^2-x^2\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = y^2 $ and $ B = x^2 $. $$ \left(y^2-x^2\right)^3 = \left( y^2 \right)^3-3 \cdot \left( y^2 \right)^2 \cdot x^2 + 3 \cdot y^2 \cdot \left( x^2 \right)^2-\left( x^2 \right)^3 = y^6-3x^2y^4+3x^4y^2-x^6 $$ |
| ② | Combine like terms: $$ -x^6+3x^4y^2-3x^2y^4+y^6 = -x^6+3x^4y^2-3x^2y^4+y^6 $$ |
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