Answer
$$(x^2-3)^3=x^6-9x^4+27x^2-27$$
Explanation
| $$ \begin{aligned}(x^2-3)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^6-9x^4+27x^2-27\end{aligned} $$ | |
| ① | Find $ \left(x^2-3\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = x^2 $ and $ B = 3 $. $$ \left(x^2-3\right)^3 = \left( x^2 \right)^3-3 \cdot \left( x^2 \right)^2 \cdot 3 + 3 \cdot x^2 \cdot 3^2-3^3 = x^6-9x^4+27x^2-27 $$ |
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