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Let us derive the cubic identities with the help of known identitites.
Expansion of $$(x+y)^3$$:
Substitute $$a = b = c = y$$ in the identity $$(x+a)(x+b)(x+c)$$ $$=$$ $$x^3$$$$+(a+b+c)x^2$$$$+(ab+bc+ca)x+abc$$.

Consider the LHS, $$(x+a)(x+b)(x+c)$$.

$$(x+a)(x+b)(x+c)$$ $$=$$ $$(x+y)(x+y)(x+y)$$

$$=$$ $$(x+y)^{3}$$

Consider the RHS, $$x^3$$$$+(a+b+c)x^2$$$$+(ab+bc+ca)x+abc$$.

$$x^3$$$$+(a+b+c)x^2$$$$+(ab+bc+ca)x+abc$$ $$=$$ $$x^3$$$$+(y+y+y)x^2$$$$+(yy+yy+yy)x+yyy$$

$$=$$ $$x^3$$$$+(3y)x^2$$$$+(y^2+y^2+y^2)x+y^3$$

$$=$$ $$x^3$$$$+3yx^2$$$$+3y^2x+y^3$$

Thus, the identity is $$(x+y)^3$$ $$=$$ $$x^3+3x^2y+3xy^2+y^3$$.

The obtained cubic identity can also be rewritten as follows:

Consider the standard identity, $$(x+y)^3$$$$=$$$$x^3+3x^2y$$$$+3xy^2+y^3$$.

Take the factor $$3xy$$ from the middle two terms of RHS.

Thus, ${\left(x+y\right)}^{3}={x}^{3}+{y}^{3}+3\mathit{xy}\left(x+y\right)$.
Expansion of $$(x-y)^3$$:
Replace $$y$$ by $$-y$$ in the cubic identity of $$(x+y)^3$$$$=$$$$x^3+3x^2y$$$$+3xy^2+y^3$$.

$$(x+(-y))^3$$ $$=$$ $$x^3+3x^2(-y)+3x(-y)^2+(-y)^3$$

$$(x-y)^3$$ $$=$$ $$x^3-3x^2y+3xy^2-y^3$$

Thus, the identity is $$(x-y)^3$$ $$=$$ $$x^3-3x^2y+3xy^2-y^3$$

The obtained cubic identity can also be rewritten as follows:

Consider the standard identity, $$(x-y)^3$$ $$=$$ $$x^3-3x^2y+3xy^2-y^3$$.

Take the factor $$3xy$$ from the middle two terms of RHS.

Thus, ${\left(x-y\right)}^{3}={x}^{3}-{y}^{3}+3\mathit{xy}\left(x-y\right)$

Let us summarize the identities...

• $$(x+y)^3$$$$=$$$$x^3+3x^2y$$$$+3xy^2+y^3$$ or ${\left(x+y\right)}^{3}={x}^{3}+{y}^{3}+3\mathit{xy}\left(x+y\right)$
• $$(x-y)^3$$ $$=$$ $$x^3-3x^2y+3xy^2-y^3$$ or ${\left(x-y\right)}^{3}={x}^{3}-{y}^{3}+3\mathit{xy}\left(x-y\right)$

Click here! to explore some examples on the expansion of cubic terms.