Question

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

Answer 1

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

Note: Here, apart from finding ${(a-b)}^3$by multiplying $(a-b)$ three times, students can directly use the formula of ${(a-b)}^3$, that is ${(a-b)}^3=a^3-b^3-3ab(a-b)$.
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, $a^3-b^3=(a-b)(a^2+b^2+ab)$
By putting the values of a and b in above equation, we get,
$\Rightarrow {\left(4\right)}^3-{\left(2\right)}^3=(4-2)({\left(4\right)}^2+{\left(2\right)}^2+4\times 2)$
By simplifying the above equation, we get,
$\begin{array}{rl}& \Rightarrow 64-8=\left(2\right)(16+4+8)\\ & \Rightarrow 56=2\left(28\right)\\ & \Rightarrow 56=56\end{array}$
Since, LHS=RHS, therefore, our formula is correct.