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Let us expand some of the cubic terms using its identities.
Write the cube in expanded form.

1. $$(2x+3y)^3$$

Let us use the identity, $$(a+b)^3$$$$=$$ $$a^3+3a^2b+3ab^2+b^3$$.

Comparing $$(2x+3y)^3$$ with $$(a+b)^3$$, we have $$a=2x$$ and $$b=3y$$.

Substitute the values in the formula.

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

$$(2x+3y)^3$$ $$=$$ $$8x^3$$$$+(2\times 4\times 3)x^2y$$$$+(3\times 2\times 9)xy^2$$$$+27y^3$$

$$= 8x^3+24x^2y+54xy^2+27y^3$$

2. $$(5x-7y)^3$$

Let us use the identity, $$(a-b)^3$$$$=$$$$a^3-3a^2b+3ab^2-b^3$$.

Comparing $$(5x-7y)^3$$ with $$(a-b)^3$$, we have $$a=5x$$ an d$$b=7y$$.

Substitute the values in the formula.

$$(5x-7y)^3$$ $$=$$ $$(5x)^3$$$$-3(5x)^2(7y)$$$$+3(5x)(7y)^2$$$$+(7y)^3$$

$$(5x-7y)^3$$ $$=$$ $$125x^3$$$$-(3\times 25\times 7)x^2y$$$$+(3\times 5 \times 49)xy^2$$$$+343y^3$$

$$(5x-7y)^3$$ $$=$$ $$125x^3$$$$-525x^2y$$$$+735xy^2$$$$+343y^3$$

4. $$(4y+5)(4y+3)(4y-7)$$

Let us use the identity, $$(x+a)(x+b)(x+c)$$ $$=$$ $$x^3+(a+b+c)x^2$$$$+(ab+bc+ca)x$$$$+abc$$

Comparing $$(4y+5)(4y+3)(4y-7)$$ with $$(x+a)(x+b)(x+c)$$, we have $$x=4y, a=5, b=3$$ an d$$c=-7$$.

Substitute the known values.

$$(4y+5)(4y+3)(4y-7)$$ $$=$$ $$(4y)^3$$$$+(5+3-7)(4y)^2$$$$+((5\times 3) + (3\times -7) +(-7\times 5))(4y)$$$$+5 \times 3 \times -7)$$

$$(4y+5)(4y+3)(4y-7)$$ $$=$$ $$64y^3+16y^2$$$$+(15-21-35)(4y)-105$$

$$(4y+5)(4y+3)(4y-7)$$ $$=$$ $$64y^3+16y^2$$$$-164y-105$$

Example:
Look for the following cases where we used the identities.

1. Expand $$(y-5)^3$$ using identity.

The above expression is in $$(a-b)^3$$ form.

We have the identity, $$(a-b)^3$$$$=$$$$a^3-3a^2b+3ab^2-b^3$$.

Substitute $$a = y$$ and $$b = 5$$ in the formula.

${\left(y-5\right)}^{3}={y}^{3}-3{\left(y\right)}^{2}\left(5\right)+3\left(y\right){\left(5\right)}^{2}-{5}^{3}$

${\left(y-5\right)}^{3}={y}^{3}-15{y}^{2}+75y-125$

2. Evaluate $$103^3$$ using identity.

$$103^3$$ $$=$$ $$(100+3)^3$$

The above expression is in $$(a+b)^3$$ form.

We have the identity, $$(a+b)^3$$ $$=$$ $$a^3+3a^2b+3ab^2+b^3$$

Substitute $$a =100$$ and $$b = 3$$ in the formula.

${\left(100+3\right)}^{2}={100}^{3}+3{\left(100\right)}^{2}\left(3\right)+3\left(100\right){\left(3\right)}^{2}+{3}^{3}$

$=1000000+\left(3×10000×3\right)+\left(3×100×9\right)+27$

$=1000000+90000+2700+27$

$=1092727$