The subtraction of $b$ cubed from $a$ cubed is written in factor form mathematically as per the difference of cubes property.

$a^3-b^3$ $\,=\,$ $(a-b)(a^2+b^2+ab)$

Let’s start to learn how to derive the $a$ cube minus $b$ cube formula algebraically by an identity.

According to the cube of difference algebraic identity, the $a$ minus $b$ whole cube can be expanded as follows.

$(a-b)^3$ $\,=\,$ $a^3$ $-$ $b^3$ $-$ $3a^2b$ $+$ $3ab^2$

The first two terms in the expansion of the $a$ minus $b$ whole cube algebraic identity represent the difference of the quantities in cube form. So, let’s focus on finding the difference of cubes in mathematical form by shifting the unnecessary terms on another side of the equation.

$\implies$ $(a-b)^3$ $+$ $3a^2b$ $-$ $3ab^2$ $\,=\,$ $a^3$ $-$ $b^3$

$\,\,\,\therefore\,\,\,\,\,\,$ $a^3$ $-$ $b^3$ $\,=\,$ $(a-b)^3$ $+$ $3a^2b$ $-$ $3ab^2$

The $a$ cubed minus $b$ cubed is expanded as a trinomial. Now, let’s try to simplify the algebraic expression by taking the common factors out from the terms.

$=\,$ $(a-b)^3$ $+$ $3a^2b$ $-$ $3ab^2$

Look at the terms on the right hand side of the equation and we can observe that three times the product of $a$ and $b$ is common in second and third terms.

$=\,$ $(a-b)^3$ $-$ $3ab \times a$ $+$ $3ab \times b$

Now, take the $3$ times product of $a$ and $b$ common from them to simplify the expression further on the right hand side of the equation.

$=\,$ $(a-b)^3$ $-$ $3ab \times (a-b)$

$=\,$ $(a-b)^3$ $-$ $3ab(a-b)$

Observe the both terms on the right-hand side of the equation. There is a factor commonly in both terms. So, it is time to take that common factor out from them to simplify the expression further.

$=\,$ $(a-b) \times (a-b)^2$ $+$ $3ab \times (a-b)$

$=\,$ $(a-b) \times \big((a-b)^2+3ab\big)$

$=\,$ $(a-b)\big((a-b)^2+3ab\big)$

The $a$ cubed minus $b$ cubed is factored as two factors. Now, observe both factors on the right hand side of the equation. There is nothing to simplify in the first factor but the first term $a$ minus $b$ whole square in the second factor can be expanded as per the square of difference algebraic identity.

$=\,$ $(a-b)(a^2+b^2-2ab+3ab)$

Now, simplify the algebraic expression in the second factor to get the factor form of the $a$ cube minus $b$ cube formula.

$\,\,\,\therefore\,\,\,\,\,\,$ $a^3$ $-$ $b^3$ $\,=\,$ $(a-b)(a^2+b^2+ab)$

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