$a^3-b^3$ $\,=\,$ $(a-b)(a^2+b^2+ab)$
Let $a$ and $b$ be two quantities in algebraic form.
As per the difference of cubes arithmetic property, the subtraction of $b$ cubed from $a$ cubed is equal to the product of the subtraction of $b$ from $a$ and the addition of sum of squares of $a$ and $b$ and the product of $a$ and $b$.
$\therefore\,\,\,$ $a^3-b^3$ $\,=\,$ $(a-b)$ $\times$ $(a^2+b^2+ab)$
Now, it is time to learn more about the $a$ cube minus $b$ cube algebraic identity.
The $a$ cube minus $b$ cube algebraic identity is alternatively written in mathematics as follows.
$x^3-y^3$ $\,=\,$ $(x-y)(x^2+y^2+xy)$
The $a$ cube minus $b$ cube algebraic identity is used in two different cases mainly.
The $a$ cube minus $b$ cube algebraic identity can be derived in two different methods.
Learn how to derive the $a$ cube minus $b$ cube formula by the algebraic identities.
Learn how to prove the $a$ cubed minus $b$ cubed identity geometrically by the volume of a cube.
Assume that $a = 5$ and $b = 2$
$(1).\,\,$ $a-b$ $\,=\,$ $5-2$ $\,=\,$ $3$
$(2).\,\,$ $a^2+b^2+ab$ $\,=\,$ $5^2+2^2+5 \times 2$ $\,=\,$ $25+4+10$ $\,=\,$ $39$
$(3).\,\,$ $a^3-b^3$ $\,=\,$ $5^3-2^3$ $\,=\,$ $125-8$ $\,=\,$ $117$
Now, calculate the product of the $a-b$ and $a^2+b^2+ab$.
$\implies$ $(a-b)$ $\times$ $(a^2+b^2+ab)$ $\,=\,$ $3 \times 39$ $\,=\,$ $117$
It is calculated that the product of them is equal to $117$, and the difference of cubes of $a$ and $b$ is also equal to $117$.
$\,\,\,\therefore\,\,\,\,\,\,$ $a^3-b^3$ $\,=\,$ $117$ $\,=\,$ $(a-b)(a^2+b^2+ab)$
You too can verify the $a$ cube minus $b$ cube algebraic identity by taking any two numbers as explained above.
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