Answer
$(x^2+x+3)(x^2-x-3)$
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: $x^4-(x^2+6x+9)=\\=(x^2)^2-((x)^2+2\cdot3\cdot x+3^2)=\\=(x^2)^2-(x+3)^2\\=(x^2+x+3)(x^2-x-3)$
