### Theory:

Product of three binomials:
Consider the product of the binomials $$(x+a)(x+b)(x+c)$$.

Apply the distributive property for the first two terms as follows:

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

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

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.

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

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

Thus we have the identity $$(x+a)(x+b)(x+c)$$ $$=$$ $$x^3$$$$+(a+b+c)x^2$$$$+(ab+bc+ca)x+abc$$.
Example:
$$(4y+5)(4y+3)(4y-7)$$

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.

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

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

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