Question

If a – b = 4 and ab = 21, find the value of ${{a}^{3}}-{{b}^{3}}$?

Answer 1

Hint: Use the identity of ${{a}^{3}}-{{b}^{3}}$ which is equal to ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ then find the value of ${{a}^{2}}+{{b}^{2}}$. Do the square of both the sides in the expression a – b = 4 and after opening ${{\left( a-b \right)}^{2}}$ expression, we will get ${{a}^{2}}+{{b}^{2}}$ as one of the terms then substitute this value of ${{a}^{2}}+{{b}^{2}}$ in ${{a}^{3}}-{{b}^{3}}$ identity.

Complete step-by-step answer:
In the below equation, we are showing the identity of ${{a}^{3}}-{{b}^{3}}$:
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.
It is given in the above question that a – b = 4 and ab = 21.
Now, we are going to calculate ${{a}^{2}}+{{b}^{2}}$:
$a-b=4$
Squaring on both the sides of the above equation we get,
$\begin{align}
  & {{\left( a-b \right)}^{2}}=16 \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}-2ab=16 \\
\end{align}$
Substituting the value of ab in the above equation, we get:
\[\begin{align}
  & {{a}^{2}}+\text{ }{{b}^{2}}-\text{ }2\left( 21 \right)\text{ }=\text{ }16 \\
 & \Rightarrow {{a}^{2}}+\text{ }{{b}^{2}}-\text{ }42\text{ }=\text{ }16 \\
 & \Rightarrow {{a}^{2}}+\text{ }{{b}^{2}}=\text{ }58 \\
\end{align}\]
Now, putting this value of ${{a}^{2}}+{{b}^{2}}$ along with a – b and ab in the identity of ${{a}^{3}}-{{b}^{3}}$ expression as follows:
$\begin{align}
  & {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right) \\
 &\Rightarrow {{a}^{3}}-{{b}^{3}}=4\left( 58+21 \right) \\
 & \Rightarrow {{a}^{3}}-{{b}^{3}}=4\left( 79 \right)=316 \\
\end{align}$
Hence the value of ${{a}^{3}}-{{b}^{3}}$ is equal to 316.

Note: The alternate way of solving the above problem as:
 In the identity of ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$, instead of solving the value of ${{a}^{2}}+{{b}^{2}}$ from expanding ${{\left( a-b \right)}^{2}}$ we can change the expression${{a}^{2}}+{{b}^{2}}$to${{\left( a-b \right)}^{2}}+2ab$. Here, the values of a – b and ab are already given in the question.
In the below steps we have worked out this alternative solution:
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
Now, solving R.H.S of the above equation as follows:
$\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
In the above expression, we can write${{a}^{2}}+{{b}^{2}}$as${{\left( a-b \right)}^{2}}+2ab$.
$\begin{align}
  & \left( a-b \right)\left( {{\left( a-b \right)}^{2}}+2ab+ab \right) \\
 & =\left( a-b \right)\left( {{\left( a-b \right)}^{2}}+3ab \right) \\
\end{align}$
Substituting the value of a – b = 4 and ab = 21 in the above expression we get,
$\begin{align}
  & =4\left( {{\left( 4 \right)}^{2}}+3\left( 21 \right) \right) \\
 & =4\left( 16+63 \right) \\
 & =4\left( 79 \right)=316 \\
\end{align}$
As we can compare the answer that we are getting from this new method with the method we have used in the solution part we have found that both the answers are the same i.e. 316.