Question

Define the formula of ${{a}^{3}}-{{b}^{3}}$.

Hint: Here we will derive the formula of $\left( {{a}^{3}}-{{b}^{3}} \right)$ by taking into account the formula of ${{\left( a-b \right)}^{3}}$which is, ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$. Then we will add $3ab\left( a-b \right)$both the sides and simplify this expression to get $\left( {{a}^{3}}-{{b}^{3}} \right)$.
Here we have to define the formula of ${{a}^{3}}-{{b}^{3}}$.
To get the formula of ${{a}^{3}}-{{b}^{3}}$, first of all, we have to derive the formula of ${{\left( a-b \right)}^{3}}$.
We know that ${{\left( a-b \right)}^{3}}=\left( a-b \right)\left( a-b \right)\left( a-b \right)$
By simplifying the RHS of above equation, we get,
\begin{align} & {{\left( a-b \right)}^{3}}=\left( a-b \right)\left( {{a}^{2}}-ab-ba+{{b}^{2}} \right) \\ & or\ {{\left( a-b \right)}^{3}}=\left( a-b \right)\left( {{a}^{2}}-2ab+{{b}^{2}} \right) \\ \end{align}
By further simplifying the RHS of above equation, we get,
${{\left( a-b \right)}^{3}}={{a}^{3}}-2{{a}^{2}}b+{{b}^{2}}a-b{{a}^{2}}+2a{{b}^{2}}-{{b}^{3}}$
Therefore we get, ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b-3{{b}^{2}}a$.
We can also write the above equation as,
${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$
Now by adding $3ab\left( a-b \right)$ on both sides of above equation, we get,
${{\left( a-b \right)}^{3}}+3ab\left( a-b \right)=\left( {{a}^{3}}-{{b}^{3}} \right)-3ab\left( a-b \right)+3ab\left( a-b \right)$
By cancelling the like terms from RHS, we get,
\begin{align} & {{\left( a-b \right)}^{3}}+3ab\left( a-b \right)=\left( {{a}^{3}}-{{b}^{3}} \right) \\ & or\ {{a}^{3}}-{{b}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right) \\ \end{align}
By taking out (a - b) common from RHS, we get,
$\Rightarrow {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left[ {{\left( a-b \right)}^{2}}+3ab \right]$
As we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, by applying it in above equation, we get,
$\Rightarrow {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}-2ab+3ab \right)$
Therefore we get, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
Hence we have found the formula for ${{a}^{3}}-{{b}^{3}}$which is equal to $\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.

Note: Here, apart from finding ${{\left( a-b \right)}^{3}}$by multiplying $\left( a-b \right)$ three times, students can directly use the formula of ${{\left( a-b \right)}^{3}}$, that is ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$.
Also students can cross check the formula by taking any value of a and b and satisfying them in formula as follows:
Let us take a = 4 and b = 2.
We have found that, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
By putting the values of a and b in above equation, we get,
$\Rightarrow {{\left( 4 \right)}^{3}}-{{\left( 2 \right)}^{3}}=\left( 4-2 \right)\left( {{\left( 4 \right)}^{2}}+{{\left( 2 \right)}^{2}}+4\times 2 \right)$
By simplifying the above equation, we get,
\begin{align} & \Rightarrow 64-8=\left( 2 \right)\left( 16+4+8 \right) \\ & \Rightarrow 56=2\left( 28 \right) \\ & \Rightarrow 56=56 \\ \end{align}
Since, LHS=RHS, therefore, our formula is correct.