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

$a$ and $b$ are literals but represents two terms. The subtraction of them is $a-b$, which is known as a binomial in algebra. The product of three same binomials is equal to the cube of that binomial. It is simply written as ${(a-b)}^3$ mathematically. The $a$ minus $b$ whole cube identity can be expanded mathematically in terms of $a$ and $b$ by the algebraic approach.

The $(a-b)^3$ can be written as two multiplying factors according to the product rule of exponents.

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

$\implies (a-b)^3$ $=$ $(a-b) \times (a-b)^2$

Expand the square of binomial by the a-b whole square formula.

$\implies (a-b)^3$ $=$ $(a-b) \times [a^2+b^2-2ab]$

Multiply both algebraic expressions by the multiplication system of algebraic expressions.

$\implies (a-b)^3$ $=$ $a \times [a^2+b^2-2ab]$ $-$ $b \times [a^2+b^2-2ab]$

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

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

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

The product of $a^2b$ is equal to the product of $ba^2$. Hence, the term $a^2b$ can be replaced by $ba^2$ and vice-versa.

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

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

The expansion of the cube of the subtraction of the terms can also be written in the following simplified form.

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

Algebraically, the proof of $a$ plus $b$ whole cube identity has proved that the cube of sum of any two terms is expanded as the sum of sum of cubes of both terms and twice the product of both terms and sum of both terms.

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