Question

Simplify:
${{\left( 2a+3b \right)}^{3}}-{{\left( 2a-3b \right)}^{3}}$

Answer 1

Hint: To simplify the given expression, we are going to use the following algebraic identity: ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right)$ . In this identity, we are going to substitute $x=\left( 2a+3b \right),y=\left( 2a-3b \right)$ and then solve the required addition and subtraction.

Complete step by step answer:
The expression given in the above problem which we have to simplify is as follows:
${{\left( 2a+3b \right)}^{3}}-{{\left( 2a-3b \right)}^{3}}$
If you carefully look at the above expression then you will find that the above expression is written in the form of the following algebraic identity which is the cube of two variables with a subtraction sign between them:
${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right)$
Now, substituting $x=\left( 2a+3b \right),y=\left( 2a-3b \right)$ in the above equation we get,
${{\left( 2a+3b \right)}^{3}}-{{\left( 2a-3b \right)}^{3}}=\left( \left( 2a+3b \right)-\left( 2a-3b \right) \right)\left( {{\left( 2a+3b \right)}^{2}}+{{\left( 2a-3b \right)}^{2}}+\left( 2a+3b \right)\left( 2a-3b \right) \right)$
Now, we are going to simplify the R.H.S of the above equation by opening all the brackets and when opening the brackets we are going to need the following algebraic identities to use:
$\begin{align}
  & {{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}; \\
 & {{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}} \\
\end{align}$
Also, we are going to use the following algebraic identity:
$\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$
Assimilating all the above algebraic identities and using them in the above equation we get,
${{\left( 2a+3b \right)}^{3}}-{{\left( 2a-3b \right)}^{3}}=\left( \left( 2a+3b \right)-\left( 2a-3b \right) \right)\left( {{\left( 2a+3b \right)}^{2}}+{{\left( 2a-3b \right)}^{2}}+\left( 2a+3b \right)\left( 2a-3b \right) \right)$
We are taking R.H.S of the above equation and then solve that we get,
$\begin{align}
  & \left( \left( 2a+3b \right)-\left( 2a-3b \right) \right)\left( {{\left( 2a+3b \right)}^{2}}+{{\left( 2a-3b \right)}^{2}}+\left( 2a+3b \right)\left( 2a-3b \right) \right) \\
 & =\left( 2a+3b-2a+3b \right)\left( {{\left( 2a \right)}^{2}}+{{\left( 3b \right)}^{2}}+2\left( 2a \right)\left( 3b \right)+{{\left( 2a \right)}^{2}}+{{\left( 3b \right)}^{2}}-2\left( 2a \right)\left( 3b \right)+{{\left( 2a \right)}^{2}}-{{\left( 3b \right)}^{2}} \right) \\
\end{align}$
As you can see that in the above expression, some positive and negative terms got cancelled out and we are left with:
$\begin{align}
  & \left( 6b \right)\left( 3{{\left( 2a \right)}^{2}}+{{\left( 3b \right)}^{2}} \right) \\
 & =\left( 6b \right)\left( 3\left( 4{{a}^{2}} \right)+9{{b}^{2}} \right) \\
 & =6b\left( 12{{a}^{2}}+9{{b}^{2}} \right) \\
\end{align}$
Multiplying $6b$ by $\left( 12{{a}^{2}}+9{{b}^{2}} \right)$ we get,
$72b{{a}^{2}}+54{{b}^{3}}$
Hence, we have simplified the given expression to$72b{{a}^{2}}+54{{b}^{3}}$.

Note: The alternate approach to the above problem is to expand ${{\left( 2a+3b \right)}^{3}}\And {{\left( 2a-3b \right)}^{3}}$ separately in ${{\left( 2a+3b \right)}^{3}}-{{\left( 2a-3b \right)}^{3}}$ and then subtract the expansion of the two cubic identities. To expand the cubic expressions we are going to use the following algebraic identity:
$\begin{align}
  & {{\left( x+y \right)}^{3}}={{\left( x \right)}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}+{{y}^{3}}; \\
 & {{\left( x-y \right)}^{3}}={{\left( x \right)}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}} \\
\end{align}$
Then substitute the values of $x=2a;y=3b$ in the above equations we get,
$\begin{align}
  & {{\left( 2a+3b \right)}^{3}}={{\left( 2a \right)}^{3}}+3{{\left( 2a \right)}^{2}}\left( 3b \right)+3\left( 2a \right){{\left( 3b \right)}^{2}}+{{\left( 3b \right)}^{3}}........(1) \\
 & {{\left( 2a-3b \right)}^{3}}={{\left( 2a \right)}^{3}}-3{{\left( 2a \right)}^{2}}\left( 3b \right)+3\left( 2a \right){{\left( 3b \right)}^{2}}-{{\left( 3b \right)}^{3}}..........(2) \\
\end{align}$
Subtracting the above two equation (2) from eq. (1) we get,
$\begin{align}
  & {{\left( 2a+3b \right)}^{3}}-{{\left( 2a-3b \right)}^{3}}=2\left( 3 \right){{\left( 2a \right)}^{2}}\left( 3b \right)+2{{\left( 3b \right)}^{3}} \\
 & \Rightarrow {{\left( 2a+3b \right)}^{3}}-{{\left( 2a-3b \right)}^{3}}=72{{a}^{2}}b+54{{b}^{3}} \\
\end{align}$
Hence, we are getting the same result of simplification as we have shown in the above solution.