Answer
$$(x-4)^2-y^2=x^2-y^2-8x+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-4)^2-y^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2-8x+16-y^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2-y^2-8x+16\end{aligned} $$ | |
| ① | Find $ \left(x-4\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 4 }$. $$ \begin{aligned}\left(x-4\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 4 + \color{red}{4^2} = x^2-8x+16\end{aligned} $$ |
| ② | Combine like terms: $$ x^2-y^2-8x+16 = x^2-y^2-8x+16 $$ |
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