Answer
$64y^2-9x^2-12x-4$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The square of a binomial sum is: $(A+B)^2=A^2+2AB+B^2$.
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $[8y+(3x+2)][8y-(3x+2)]=(8y)^2-(3x+2)^2=64y^2-(3x)^2-2\cdot3x \cdot2-2^2=64y^2-9x^2-12x-4$
