Answer
$x^4y^4-6x^2y^2+9$
Work Step by Step
$$(x^2y^2-3)^2$$ To square a binomial, multiply it by itself: $$=(x^2y^2-3)(x^2y^2-3)$$ To multiply two binomials (expressions with two terms) multiply each term in the first binomial by each term in the second binomial:$$=(x^2y^2]times x^2y^2)+[x^2y^2\times (-3)]+(-3\times x^2y^2)+[-3\times (-3)]$$ $$=x^4y^4-3x^2y^2-3x^2y^2+9$$ $$=x^4y^4-6x^2y^2+9$$
