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We will now extend this idea of expanding algebraic terms/numbers with cubic order using the identities.
Let us derive the cubic identities with the help of known square identities.
1. We will now prove $$(a+b)^3$$$$=$$ $$a^3+3a^2b+3ab^2+b^3$$ by direct multiplication.

Consider the LHS $$(a+b)^3$$.

Here $$(a+b)$$ raised to the power $$3$$.  It means we need to multiply $$(a+b)$$ by itself for two times.

That is $$(a+b)\times(a+b)\times(a+b)$$ $$=$$ $$(a+b)^3$$

$$(a+b)^3$$ $$=$$ $$[(a+b)\times(a+b)]$$$$\times(a+b)$$

$$=$$ $$(a+b)^2\times(a+b)$$

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

Apply the distributive property.

$$=$$ $$a^2\times a$$$$+b^2\times a$$$$+2ab\times a$$$$+a^2\times b$$$$+b^2\times b$$$$+2ab\times b$$

$$=$$ $$a^3+ab^2+2a^2b+a^2b+b^3+2ab^2$$

$$=$$ $$a^3+3a^2b+3ab^2+b^3$$

$$=$$ RHS

Thus, the identity is $$(a+b)^3$$ $$=$$ $$a^3+3a^2b+3ab^2+b^3$$.

2. We will now prove $$(a-b)^3$$ $$=$$ $$a^3-3a^2b+3ab^2-b^3$$ by direct multiplication.

Consider the LHS $$(a-b)^3$$.

Here $$(a-b)$$ raised to the power $$3$$.  It means we need to multiply $$(a-b)$$ by itself for two times.

That is $$(a-b)\times (a-b)\times(a-b)$$$$=$$$$(a-b)^3$$.

$$(a-b)^3$$ $$=$$ $$[(a-b)\times(a-b)]$$$$\times(a-b)$$

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

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

Apply the distributive property.

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

$$=$$ $$a^3+ab^2-2a^2b-a^2b-b^3+2ab^2$$

$$=$$ $$a^3-3a^2b+3ab^2-b^3$$

$$=$$ RHS

Thus, the identity is $$(a-b)^3$$ $$=$$ $$a^3-3a^2b+3ab^2-b^3$$

3. We will now prove $$(x+a)(x+b)(x+c)$$ $$=$$ $$x^3$$$$+(a+b+c)x^2$$$$+(ab+bc+ca)x+abc$$.

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

Apply the distributive property for the first two terms.

$$[(x+a)(x+b)](x+c)$$ $$=$$ $$[(x\times x)]$$$$+(x\times b)$$$$+(a\times x)$$$$+(a\times b)]$$$$(x+c)$$

$$=$$ $$(x^2+bx+ax+ab)$$$$(x+c)$$

Again apply distributive law.

$$(x^2+bx+ax+ab)(x+c)$$

$$=$$ $$(x^2\times x)$$$$+(bx\times x)$$$$+(ax\times x)$$$$+(ab\times x)$$$$+(x^2\times c)$$$$+(bx\times c)$$$$+(ax\times c)$$$$+(ab\times c)$$.

$$=$$ $$x^3+bx^2$$$$+ax^2+abx$$$$+cx^2+bcx$$$$+acx+abc$$

Separate the cubic, square, variables and constants terms.

$$=$$ $$x^3+ax^2$$$$+bx^2+cx^2$$$$+abx+bcx$$$$+acx+abc$$

$$=$$ $$x^3+(a+b+c)x^2+(ab+bc+ac)x+abc$$

$$=$$ RHS

Thus we have the identity $$(x+a)(x+b)(x+c)$$ $$=$$ $$x^3$$$$+(a+b+c)x^2$$$$+(ab+bc+ca)x+abc$$.
Let us summarize identities.
${I.\phantom{\rule{0.147em}{0ex}}\left(a+b\right)}^{3}={a}^{3}+3{a}^{2}b+\mathit{3a}{b}^{2}+{b}^{3}\phantom{\rule{0.147em}{0ex}}$

$\mathit{II}.{\phantom{\rule{0.147em}{0ex}}\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+\mathit{3a}{b}^{2}-{b}^{3}\phantom{\rule{0.147em}{0ex}}$

$\mathit{III}.\phantom{\rule{0.147em}{0ex}}\left(x+a\right)\left(x+b\right)\left(x+c\right)={x}^{3}+\left(a+b+c\right){x}^{2}+\left(\mathit{ab}+\mathit{bc}+\mathit{ca}\right)x+\mathit{abc}$