Consider $\left(4x-3\right)\left(4x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.
$$\left(4x\right)^{2}-9+\left(5x-4\right)^{2}$$
Expand $\left(4x\right)^{2}$.
$$4^{2}x^{2}-9+\left(5x-4\right)^{2}$$
Calculate $4$ to the power of $2$ and get $16$.
$$16x^{2}-9+\left(5x-4\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(5x-4\right)^{2}$.
Consider $\left(4x-3\right)\left(4x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.
$$\left(4x\right)^{2}-9+\left(5x-4\right)^{2}$$
Expand $\left(4x\right)^{2}$.
$$4^{2}x^{2}-9+\left(5x-4\right)^{2}$$
Calculate $4$ to the power of $2$ and get $16$.
$$16x^{2}-9+\left(5x-4\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(5x-4\right)^{2}$.
