Question

Simplify the following:
(i) ${{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}$
(ii) ${{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}$
(iii) ${{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 2.5p-1.5q \right)}^{2}}$
(iv) ${{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c$

Answer 1

Hint: First, before proceeding for this, we must know the following formulas that is used in all the given expressions and helps us to solve them individually as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. Then, by using the above stated rules, we can find the value of all the expressions given in the parts of the question and get the desired result for each part.

Complete step by step answer:
In this question, we are supposed to simplify the given expressions to get the desired answers.
So, before proceeding for this, we must know the following formulas that is used in all the given expressions and helps us to solve them individually as:
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align}$
So, starting with the first one, we have ${{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}$and for solving it we have to use the property stated above as ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, then we get:
$\begin{align}
  & {{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}={{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{b}^{2}} \right)}^{2}}-2\left( {{a}^{2}} \right)\left( {{b}^{2}} \right) \\
 & \Rightarrow {{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}={{a}^{4}}+{{b}^{4}}-2{{a}^{2}}{{b}^{2}} \\
\end{align}$
So, we get the simplified result for ${{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}$as ${{a}^{4}}+{{b}^{4}}-2{{a}^{2}}{{b}^{2}}$.
Then, moving towards second one, we have ${{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}$where we need to use both the above stated formulas as ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, then we get:
$\begin{align}
  & {{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}={{\left( 7m \right)}^{2}}+{{\left( 8n \right)}^{2}}-2\left( 7m \right)\left( 8n \right)+{{\left( 7m \right)}^{2}}+{{\left( 8n \right)}^{2}}+2\left( 7m \right)\left( 8n \right) \\
 & \Rightarrow {{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}=49{{m}^{2}}+64{{n}^{2}}+49{{m}^{2}}+64{{n}^{2}} \\
 & \Rightarrow {{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}=98{{m}^{2}}+128{{n}^{2}} \\
\end{align}$
So, we get the simplified result for ${{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}$as $98{{m}^{2}}+128{{n}^{2}}$.
Then, moving towards third one, we have ${{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 2.5p-1.5q \right)}^{2}}$where we need to use both the above stated formulas as ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, then we get:
\[\begin{align}
  & {{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 2.5p-1.5q \right)}^{2}}={{\left( 2.5p \right)}^{2}}+{{\left( 1.5q \right)}^{2}}-2\left( 2.5p \right)\left( 1.5q \right)-\left[ {{\left( 2.5p \right)}^{2}}+{{\left( 1.5q \right)}^{2}}-2\left( 2.5p \right)\left( 1.5q \right) \right] \\
 & \Rightarrow {{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 2.5p-1.5q \right)}^{2}}={{\left( 2.5p \right)}^{2}}+{{\left( 1.5q \right)}^{2}}-2\left( 2.5p \right)\left( 1.5q \right)-{{\left( 2.5p \right)}^{2}}-{{\left( 1.5q \right)}^{2}}+2\left( 2.5p \right)\left( 1.5q \right) \\
 & \Rightarrow {{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 2.5p-1.5q \right)}^{2}}=0 \\
\end{align}\]So, we get the simplified result for ${{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 2.5p-1.5q \right)}^{2}}$as 0.
Then, moving towards fourth one, we have ${{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c$where we need to use the above stated formula as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, then we get:
$\begin{align}
  & {{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c={{\left( ab \right)}^{2}}+{{\left( bc \right)}^{2}}+2\left( ab \right)\left( bc \right)-2a{{b}^{2}}c \\
 & \Rightarrow {{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+2a{{b}^{2}}c-2a{{b}^{2}}c \\
 & \Rightarrow {{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}} \\
\end{align}$
So, we get the simplified result for ${{\left( ab+bc \right)}^{2}}-2a{{b}^{2}}c$as ${{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}$.

Note:
Now, to solve these types of the questions we need to know some of the basic formulas for the simplification of the to variables identities. So, some of them are as follows:
$\begin{align}
  & {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\
 & {{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}={{\left( a+b \right)}^{2}}{{\left( a-b \right)}^{2}} \\
\end{align}$
So, it gives an alternative approach to solve the first part of the question and we will get the same answer.