Question
Answer
$$-2(-3x^2+y^2+z^2)(x^2+y^2+z^2)=6x^4+4x^2y^2+4x^2z^2-2y^4-4y^2z^2-2z^4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-2(-3x^2+y^2+z^2)(x^2+y^2+z^2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(6x^2-2y^2-2z^2)(x^2+y^2+z^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}6x^4+4x^2y^2+4x^2z^2-2y^4-4y^2z^2-2z^4\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{-2} $ by $ \left( -3x^2+y^2+z^2\right) $ $$ \color{blue}{-2} \cdot \left( -3x^2+y^2+z^2\right) = 6x^2-2y^2-2z^2 $$ |
| ② | Multiply each term of $ \left( \color{blue}{6x^2-2y^2-2z^2}\right) $ by each term in $ \left( x^2+y^2+z^2\right) $. $$ \left( \color{blue}{6x^2-2y^2-2z^2}\right) \cdot \left( x^2+y^2+z^2\right) = \\ = 6x^4+6x^2y^2+6x^2z^2-2x^2y^2-2y^4-2y^2z^2-2x^2z^2-2y^2z^2-2z^4 $$ |
| ③ | Combine like terms: $$ 6x^4+ \color{blue}{6x^2y^2} + \color{red}{6x^2z^2} \color{blue}{-2x^2y^2} -2y^4 \color{green}{-2y^2z^2} \color{red}{-2x^2z^2} \color{green}{-2y^2z^2} -2z^4 = \\ = 6x^4+ \color{blue}{4x^2y^2} + \color{red}{4x^2z^2} -2y^4 \color{green}{-4y^2z^2} -2z^4 $$ |
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