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) $${{\left( {{m}^{2}}-{{n}^{2}}m \right)}^{2}}+2{{m}^{3}}{{n}^{2}}$$

Question

Vedantu Content Team · Accepted Answer

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.

Evaluate the following : -(i) (a2−b2)2(ii) (2x+5)2−(2x−5)2(iii) (7m−8n)2+(7m+8n)2(iv) (4m+5n)2+(4n+5m)2(v) (2.5p−1.5q)2−(1.5p−2.5q)2(vi) (ab+bc)2−2ab2c(vii) (m2−n2m)2+2m3n2