$x^3-y^3$ $\,=\,$ $(x-y)(x^2+y^2+xy)$
Let’s assume, $x$ and $y$ denote two quantities in algebraic form.
According to the difference of cubes arithmetic property, the subtraction of $y$ cube from $x$ cube is equal to the product of the subtraction of $y$ from $x$ and the addition of the sum of squares of $x$ and $y$ and the product of $x$ and $y$.
$\therefore\,\,\,$ $x^3-y^3$ $\,=\,$ $(x-y)$ $\times$ $(x^2+y^2+xy)$
Now, let us learn much more about the $x$ cube minus $y$ cube algebraic identity.
The $x$ cube minus $y$ cube algebraic identity is alternatively written in the following mathematical form.
$a^3-b^3$ $\,=\,$ $(a-b)(a^2+b^2+ab)$
The $x$ cubed minus $y$ cubed algebraic identity is mainly used in two different cases.
The $x$ cube minus $y$ cube algebraic identity can be derived mathematically in two methods.
Learn how to prove the $x$ cubed minus $y$ cubed identity by the algebraic identities.
Learn how to derive the $x$ cube minus $y$ cube formula geometrically by the volume of a cube.
Assume that $x = 3$ and $y = 1$
$(1).\,\,$ $x-y$ $\,=\,$ $3-1$ $\,=\,$ $2$
$(2).\,\,$ $x^2+y^2+xy$ $\,=\,$ $3^2+1^2+3 \times 1$ $\,=\,$ $9+1+3$ $\,=\,$ $13$
$(3).\,\,$ $x^3-y^3$ $\,=\,$ $3^3-1^3$ $\,=\,$ $27-1$ $\,=\,$ $26$
Now, find the product of the $x-y$ and $x^2+y^2+xy$.
$\implies$ $(x-y)$ $\times$ $(x^2+y^2+xy)$ $\,=\,$ $2 \times 13$ $\,=\,$ $26$
It is evaluated that the product of them is equal to $26$, and the difference of cubes of $x$ and $y$ is also equal to $26$.
$\,\,\,\therefore\,\,\,\,\,\,$ $x^3-y^3$ $\,=\,$ $26$ $\,=\,$ $(x-y)(x^2+y^2+xy)$
In this way, you can verify the $x$ cube minus $y$ cube algebraic identity by taking any two numbers.
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