# Proof of $(a-b)^3$ formula in Algebraic Method

The expansion for the cube of a difference of two terms is expressed in the following mathematical form in algebraic mathematics.

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

$a$ 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.

### Product form of the Binomials

Let’s write the cube of the difference of two terms in the following form for our convenience.

$\implies$ $(a-b)^3$ $\,=\,$ $(a-b)^{2\,+\,1}$

The cube of a difference basis binomial can be split into multiplying factors by using the product rule of exponents.

$\implies$ $(a-b)^3$ $\,=\,$ $(a-b)^2 \times (a-b)^1$

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

### Expanding the square of Binomial

The square of difference of two terms can be expanded at the right hand side of the equation by the a minus b whole square algebraic identity.

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

### Multiplying the Algebraic Expressions

In the above step, the difference of two terms whole cubed is written as the product of two algebraic expressions. It expresses that the cube of difference of terms can be expanded by multiplying the algebraic expressions. Therefore, let’s multiply them to find the expansion for the $a$ minus $b$ whole cube algebraic identity.

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

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

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

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

According to the commutative property, 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.

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

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

The cube of difference of terms is expanded as an algebraic expression which consists of four terms. Thus, the $a$ minus $b$ whole cube algebraic identity is proved in algebraic method by multiplying the algebraic expressions.

### Simplifying the expansion

The quadrinomial can be simplified further by taking the common factors out from the terms.

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

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

Therefore, it is proved that the $a$ minus $b$ whole cube is equal to the $a$ cubed minus $b$ cubed minus $3ab$ times $a$ minus $b$. In this way, the expansion of the cube of difference is proved in algebraic mathematics.

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