Answer
$$(x-3)^3-8=x^3-9x^2+27x-35$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-3)^3-8& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^3-9x^2+27x-27-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^3-9x^2+27x-35\end{aligned} $$ | |
| ① | Find $ \left(x-3\right)^3 $ using formula $$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$where $ A = x $ and $ B = 3 $. $$ \left(x-3\right)^3 = x^3-3 \cdot x^2 \cdot 3 + 3 \cdot x \cdot 3^2-3^3 = x^3-9x^2+27x-27 $$ |
| ② | Combine like terms: $$ x^3-9x^2+27x \color{blue}{-27} \color{blue}{-8} = x^3-9x^2+27x \color{blue}{-35} $$ |
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