Sure, let's solve each of the given binomial products step by step:
### (i) [tex]\((2x + y)(2x + y)\)[/tex]
This is a perfect square trinomial:
[tex]\[
(2x + y)^2 = (2x)^2 + 2(2x)(y) + y^2 = 4x^2 + 4xy + y^2
\][/tex]
So, the product is [tex]\(\boxed{4x^2 + 4xy + y^2}\)[/tex].
### (ii) [tex]\((a + 2b)(a - 2b)\)[/tex]
This is a difference of squares:
[tex]\[
(a + 2b)(a - 2b) = a^2 - (2b)^2 = a^2 - 4b^2
\][/tex]
So, the product is [tex]\(\boxed{a^2 - 4b^2}\)[/tex].
### (iii) [tex]\(\left(a^2 + bc\right)\left(a^2 - bc\right)\)[/tex]
This is also a difference of squares:
[tex]\[
(a^2 + bc)(a^2 - bc) = (a^2)^2 - (bc)^2 = a^4 - b^2c^2
\][/tex]
So, the product is [tex]\(\boxed{a^4 - b^2c^2}\)[/tex].
### (iv) [tex]\(\left(\frac{4x}{5} - \frac{3y}{4}\right)\left(\frac{4x}{5} + \frac{3y}{4}\right)\)[/tex]
This is another difference of squares:
[tex]\[
\left(\frac{4x}{5} - \frac{3y}{4}\right)\left(\frac{4x}{5} + \frac{3y}{4}\right) = \left(\frac{4x}{5}\right)^2 - \left(\frac{3y}{4}\right)^2 = \frac{16x^2}{25} - \frac{9y^2}{16}
\][/tex]
So, the product is [tex]\(\boxed{\frac{16x^2}{25} - \frac{9y^2}{16}}\)[/tex].
### (v) [tex]\(\left(2x + \frac{3}{y}\right)\left(2x - \frac{3}{y}\right)\)[/tex]
This is again a difference of squares:
[tex]\[
\left(2x + \frac{3}{y}\right)\left(2x - \frac{3}{y}\right) = (2x)^2 - \left(\frac{3}{y}\right)^2 = 4x^2 - \frac{9}{y^2}
\][/tex]
So, the product is [tex]\(\boxed{4x^2 - \frac{9}{y^2}}\)[/tex].
### (vi) [tex]\(\left(2a^3 + b^3\right)\left(2a^3 - b^3\right)\)[/tex]
This is a difference of squares:
[tex]\[
\left(2a^3 + b^3\right)\left(2a^3 - b^3\right) = (2a^3)^2 - (b^3)^2 = 4a^6 - b^6
\][/tex]
So, the product is [tex]\(\boxed{4a^6 - b^6}\)[/tex].
### (vii) [tex]\(\left(x^4 + \frac{2}{x^2}\right)\left(x^4 - \frac{2}{x^2}\right)\)[/tex]
This is also a difference of squares:
[tex]\[
\left(x^4 + \frac{2}{x^2}\right)\left(x^4 - \frac{2}{x^2}\right) = (x^4)^2 - \left(\frac{2}{x^2}\right)^2 = x^8 - \frac{4}{x^4}
\][/tex]
So, the product is [tex]\(\boxed{x^8 - \frac{4}{x^4}}\)[/tex].
### (viii) [tex]\(\left(x^3 + \frac{1}{x^3}\right)\left(x^3 - \frac{1}{x^3}\right)\)[/tex]
This is once again a difference of squares:
[tex]\[
\left(x^3 + \frac{1}{x^3}\right)\left(x^3 - \frac{1}{x^3}\right) = (x^3)^2 - \left(\frac{1}{x^3}\right)^2 = x^6 - \frac{1}{x^6}
\][/tex]
So, the product is [tex]\(\boxed{x^6 - \frac{1}{x^6}}\)[/tex].
These are the final expanded forms for each of the given binomials.
