Answer:
[tex]\left(2g^4+3a^2\right)\left(4g^8-6g^4a^2+9a^4\right)[/tex]
Step-by-step explanation:
[tex]27a^6+8g^{12}[/tex]
[tex]\mathrm{Rewrite\:}27a^6+8g^{12}\mathrm{\:as\:}\left(3a^2\right)^3+\left(2g^4\right)^3[/tex]
[tex]\mathrm{Apply\:Sum\:of\:Cubes\:Formula:\:}x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)[/tex]
[tex]\left(3a^2\right)^3+\left(2g^4\right)^3=\left(3a^2+2g^4\right)\left(3^2a^4-3\cdot \:2g^4a^2+2^2g^8\right)[/tex]
[tex]=\left(2g^4+3a^2\right)\left(2^2g^8-3\cdot \:2g^4a^2+3^2a^4\right)[/tex]
[tex]=\left(2g^4+3a^2\right)\left(4g^8-6g^4a^2+9a^4\right)[/tex]
Answer:
b
Step-by-step explanation:
correct on edge 2021
the other one isnt even an option
