Evaluate the following : -
(i) \[{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}\]
(ii) \[{{\left( 2x+5 \right)}^{2}}-{{\left( 2x-5 \right)}^{2}}\]
(iii) \[{{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}\]
(iv) \[{{\left( 4m+5n \right)}^{2}}+{{\left( 4n+5m \right)}^{2}}\]
(v) \[{{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 1.5p-2.5q \right)}^{2}}\]
(vi) \[{{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c\]
(vii) \[{{\left( {{m}^{2}}-{{n}^{2}}m \right)}^{2}}+2{{m}^{3}}{{n}^{2}}\]

Hint: Use the formulae: -
\[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
 & \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) \\
\end{align}\]

Complete step-by-step answer:
(i) \[{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}\]
We know, \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)\]
And \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
\[\begin{align}
  & {{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}={{\left( {{a}^{2}} \right)}^{2}}-2{{a}^{2}}{{b}^{2}}+{{\left( {{b}^{2}} \right)}^{2}} \\
 & {{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}={{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}} \\
 & {{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}=\left( {{a}^{4}}+{{b}^{4}} \right)-2{{a}^{2}}{{b}^{2}} \\
\end{align}\]
(ii) \[{{\left( 2x+5 \right)}^{2}}-{{\left( 2x-5 \right)}^{2}}\]
Use the formulae to get the required simplification,
\[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
\end{align}\]
Here a=2x and b=5.
\[\begin{align}
  & {{\left( 2x+5 \right)}^{2}}-{{\left( 2x-5 \right)}^{2}}=\left[ {{\left( 2x \right)}^{2}}+2\times \left( 2x \right)\times 5+{{\left( 5 \right)}^{2}} \right]-\left[ {{\left( 2x \right)}^{2}}-2\times \left( 2x \right)\times 5+{{\left( 5 \right)}^{2}} \right] \\
 & {{\left( 2x+5 \right)}^{2}}-{{\left( 2x-5 \right)}^{2}}=\left( 4{{x}^{2}}+20x+25 \right)-\left( 4{{x}^{2}}-20x+25 \right) \\
\end{align}\]
Opening the bracket and simplifying it,
\[\begin{align}
  & =4{{x}^{2}}+20x+25-4{{x}^{2}}+20x-25 \\
 & =20x+20x=40x \\
 & \therefore {{\left( 2x+5 \right)}^{2}}-{{\left( 2x-5 \right)}^{2}}=40x \\
\end{align}\]
(iii) \[{{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}\]
Use the formula to get the required simplification,
\[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
\end{align}\]
Here a=7m and b=8n.
\[\begin{align}
  & {{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}=\left[ {{\left( 7m \right)}^{2}}+2\times \left( 7m \right)\times \left( 8n \right)+{{\left( 8n \right)}^{2}} \right]+\left[ {{\left( 7m \right)}^{2}}-2\times \left( 7m \right)\times \left( 8n \right)+{{\left( 8n \right)}^{2}} \right] \\
 & {{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}=\left( 49{{m}^{2}}-112mn+64{{n}^{2}} \right)+\left( 49{{m}^{2}}+112mn+64{{n}^{2}} \right) \\
\end{align}\]
Opening the bracket and simplifying it,
\[\begin{align}
  & =49{{m}^{2}}-112mn+64{{n}^{2}}+49{{m}^{2}}+112mn+64{{n}^{2}} \\
 & =98{{m}^{2}}+128{{n}^{2}} \\
 & =2\left( 49{{m}^{2}}+64{{n}^{2}} \right) \\
 & \therefore {{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}=2\left( 49{{m}^{2}}+64{{n}^{2}} \right) \\
\end{align}\]
(iv) \[{{\left( 4m+5n \right)}^{2}}+{{\left( 4n+5m \right)}^{2}}\]
We can use the formula, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[\begin{align}
  & {{\left( 4m+5n \right)}^{2}}+{{\left( 4n+5m \right)}^{2}} \\
 & =\left[ {{\left( 4m \right)}^{2}}+2\left( 4m \right)\left( 5n \right)+{{\left( 5n \right)}^{2}} \right]+\left[ {{\left( 4n \right)}^{2}}+2\left( 4n \right)\left( 5m \right)+{{\left( 5m \right)}^{2}} \right] \\
 & =\left( 16{{m}^{2}}+40mn+25{{n}^{2}} \right)+\left( 16{{n}^{2}}+40mn+25{{m}^{2}} \right) \\
\end{align}\]
Opening the bracket and simplifying it,
\[\begin{align}
  & =16{{m}^{2}}+40mn+25{{n}^{2}}+16{{n}^{2}}+40mn+25{{m}^{2}} \\
 & =\left( 16{{m}^{2}}+25{{m}^{2}} \right)+\left( 40mn+40mn \right)+\left( 25{{n}^{2}}+16{{n}^{2}} \right) \\
 & =41{{m}^{2}}+80mn+41{{n}^{2}} \\
 & \therefore {{\left( 4m+5n \right)}^{2}}+{{\left( 4n+5m \right)}^{2}}=41{{m}^{2}}+80mn+41{{n}^{2}} \\
\end{align}\]
(v) \[{{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 1.5p-2.5q \right)}^{2}}\]
We can use the formula, \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
\[\begin{align}
  & {{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 1.5p-2.5q \right)}^{2}}= \\
 & =\left[ {{\left( 2.5p \right)}^{2}}-2\left( 2.5p \right)\left( 1.5q \right)+{{\left( 1.5q \right)}^{2}} \right]-\left[ {{\left( 1.5p \right)}^{2}}-2\left( 1.5p \right)\left( 2.5q \right)+{{\left( 2.5q \right)}^{2}} \right] \\
 & =\left( 6.25{{p}^{2}}-7.5pq+2.25{{q}^{2}} \right)-\left( 2.25{{p}^{2}}-7.5pq+6.25{{q}^{2}} \right) \\
 & =6.25{{p}^{2}}-7.5pq+2.25{{q}^{2}}-2.25{{p}^{2}}+7.5pq-6.25{{q}^{2}} \\
\end{align}\]
Opening the bracket and simplifying it,
\[\begin{align}
  & =\left( 6.25-2.25 \right){{p}^{2}}+\left( 2.25-6.25 \right){{q}^{2}} \\
 & =4{{p}^{2}}+\left( -4 \right){{q}^{2}}=4\left( {{p}^{2}}-{{q}^{2}} \right)=4\left( p+q \right)\left( p-q \right) \\
 & \therefore {{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 1.5p-2.5q \right)}^{2}}=4\left( {{p}^{2}}-{{q}^{2}} \right) \\
\end{align}\]
(vi) \[{{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c\]
Using the formula, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[{{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c={{\left( ab \right)}^{2}}+2\left( ab \right)\left( bc \right)+{{\left( bc \right)}^{2}}-2a{{b}^{2}}c\]
\[{{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c={{a}^{2}}{{b}^{2}}+2a{{b}^{2}}c+{{b}^{2}}{{c}^{2}}-2a{{b}^{2}}c\]
\[\begin{align}
  & {{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}={{b}^{2}}\left( {{a}^{2}}+{{c}^{2}} \right) \\
 & \therefore {{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c={{b}^{2}}\left( {{a}^{2}}+{{c}^{2}} \right) \\
\end{align}\]
(vii) \[{{\left( {{m}^{2}}-{{n}^{2}}m \right)}^{2}}+2{{m}^{3}}{{n}^{2}}\]
Using the formula, \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
\[\begin{align}
  & {{\left( {{m}^{2}}-{{n}^{2}}m \right)}^{2}}+2{{m}^{3}}{{n}^{2}}={{\left( {{m}^{2}} \right)}^{2}}-2\left( {{m}^{2}} \right)\left( {{n}^{2}}m \right)+{{\left( {{n}^{2}}m \right)}^{2}}+2{{m}^{3}}{{n}^{2}} \\
 & ={{m}^{4}}-2{{n}^{2}}{{m}^{3}}+{{n}^{4}}{{m}^{2}}+2{{n}^{2}}{{m}^{3}} \\
 & ={{m}^{4}}+{{n}^{4}}{{m}^{2}} \\
 & ={{m}^{2}}\left( {{m}^{2}}+{{n}^{4}} \right) \\
 & \therefore {{\left( {{m}^{2}}-{{n}^{2}}m \right)}^{2}}+2{{m}^{3}}{{n}^{2}}={{m}^{2}}\left( {{m}^{2}}+{{n}^{4}} \right) \\
\end{align}\]

Note: Be cautious while simplifying it so you don’t misplace the sign and also the variables. Misplacing them might change the entire simplification.