Answer
$$m^2(f^2-2fz+z^2+y^2)-(fm-nz)^2=-2fm^2z+2fmnz+m^2y^2+m^2z^2-n^2z^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}m^2(f^2-2fz+z^2+y^2)-(fm-nz)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}m^2(f^2-2fz+z^2+y^2)-(1f^2m^2-2fmnz+n^2z^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}f^2m^2-2fm^2z+m^2z^2+m^2y^2-(1f^2m^2-2fmnz+n^2z^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}f^2m^2-2fm^2z+m^2z^2+m^2y^2-f^2m^2+2fmnz-n^2z^2 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{f^2m^2}-2fm^2z+m^2z^2+m^2y^2 -\cancel{f^2m^2}+2fmnz-n^2z^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-2fm^2z+2fmnz+m^2y^2+m^2z^2-n^2z^2\end{aligned} $$ | |
| ① | Find $ \left(fm-nz\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ fm } $ and $ B = \color{red}{ nz }$. $$ \begin{aligned}\left(fm-nz\right)^2 = \color{blue}{\left( fm \right)^2} -2 \cdot fm \cdot nz + \color{red}{\left( nz \right)^2} = f^2m^2-2fmnz+n^2z^2\end{aligned} $$ |
| ② | Multiply $ \color{blue}{m^2} $ by $ \left( f^2-2fz+z^2+y^2\right) $ $$ \color{blue}{m^2} \cdot \left( f^2-2fz+z^2+y^2\right) = f^2m^2-2fm^2z+m^2z^2+m^2y^2 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( f^2m^2-2fmnz+n^2z^2 \right) = -f^2m^2+2fmnz-n^2z^2 $$ |
| ④ | Combine like terms: $$ \, \color{blue}{ \cancel{f^2m^2}} \,-2fm^2z+m^2z^2+m^2y^2 \, \color{blue}{ -\cancel{f^2m^2}} \,+2fmnz-n^2z^2 = -2fm^2z+2fmnz+m^2y^2+m^2z^2-n^2z^2 $$ |
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