Find the product of the following binomials:

Solution:

(i) We will use the identity $(a+b)^{2}=a^{2}+2 a b+b^{2}$ in the given expression to find the product.

$(2 x+y)(2 x+y)$

$=(2 x+y)^{2}$

$=(2 x)^{2}+2(2 x)(y)+y^{2}$

$=4 x^{2}+4 x y+y^{2}$

(ii) We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ in the given expression to find the product.

$(a+2 b)(a-2 b)$

$=a^{2}-(2 b)^{2}$

$=a^{2}-4 b^{2}$

(iii) We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ in the given expression to find the product.

$\left(a^{2}+b c\right)\left(a^{2}-b c\right)$

$=\left(a^{2}\right)^{2}-(b c)^{2}$

$=a^{4}-b^{2} c^{2}$

(iv)We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ in the given expression to find the product.

$\left(\frac{4 x}{5}-\frac{3 y}{4}\right)\left(\frac{4 x}{5}+\frac{3 y}{4}\right)$

$=\left(\frac{4 x}{5}\right)^{2}-\left(\frac{3 y}{4}\right)^{2}$

$=\frac{16 x^{2}}{25}-\frac{9 y^{2}}{16}$

(v) We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ in the given expression to find the product.

$\left(2 x+\frac{3}{y}\right)\left(2 x-\frac{3}{y}\right)$

$=(2 x)^{2}-\left(\frac{3}{y}\right)^{2}$

$=4 x^{2}-\frac{9}{y^{2}}$

(vi) We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ in the given expression to find the product.

$\left(2 a^{3}+b^{3}\right)\left(2 a^{3}-b^{3}\right)$

$=\left(2 a^{3}\right)^{2}-\left(b^{3}\right)^{2}$

$=4 a^{6}-b^{6}$

(vii) We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ in the given expression to find the product.

$\left(x^{4}+\frac{2}{x^{2}}\right)\left(x^{4}-\frac{2}{x^{2}}\right)$

$=\left(x^{4}\right)^{2}-\left(\frac{2}{x^{2}}\right)^{2}$

$=x^{8}-\frac{4}{x^{4}}$

(viii) We will use the identity $(a+b)(a-b)=a^{2}-b^{2}$ in the given expression to find the product.

$\left(x^{3}+\frac{1}{x^{3}}\right)\left(x^{3}-\frac{1}{x^{3}}\right)$

$=\left(x^{3}\right)^{2}-\left(\frac{1}{x^{3}}\right)^{2}$

$=x^{6}-\frac{1}{x^{6}}$