what is the factored form of $27d^{6}+8g^{12}$?\n$(3d^{2}+2g^{4})(3d^{2}-6d^{2}g^{4}+2g^{4})$\n$(3d^{2}+2g^{4…

what is the factored form of $27d^{6}+8g^{12}$?\n$(3d^{2}+2g^{4})(3d^{2}-6d^{2}g^{4}+2g^{4})$\n$(3d^{2}+2g^{4})(9d^{4}-6d^{2}g^{4}+4g^{8})$\n$(3d^{2}+2g^{4})(9d^{4}+6d^{2}g^{4}+4g^{8})$\n$(3d^{2}+2g^{4})(3d^{4}-6d^{2}g^{4}+2g^{8})

what is the factored form of $27d^{6}+8g^{12}$?\n$(3d^{2}+2g^{4})(3d^{2}-6d^{2}g^{4}+2g^{4})$\n$(3d^{2}+2g^{4})(9d^{4}-6d^{2}g^{4}+4g^{8})$\n$(3d^{2}+2g^{4})(9d^{4}+6d^{2}g^{4}+4g^{8})$\n$(3d^{2}+2g^{4})(3d^{4}-6d^{2}g^{4}+2g^{8})

Answer

Explanation:

Step1: Recall sum - of - cubes formula

The sum - of - cubes formula is $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. We can rewrite $27d^{6}+8g^{12}$ as $(3d^{2})^{3}+(2g^{4})^{3}$. Here, $a = 3d^{2}$ and $b=2g^{4}$.

Step2: Apply the sum - of - cubes formula

Substitute $a = 3d^{2}$ and $b = 2g^{4}$ into the formula $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. We get $(3d^{2}+2g^{4})((3d^{2})^{2}-(3d^{2})(2g^{4})+(2g^{4})^{2})$.

Step3: Simplify the second factor

$(3d^{2})^{2}=9d^{4}$, $(3d^{2})(2g^{4}) = 6d^{2}g^{4}$, and $(2g^{4})^{2}=4g^{8}$. So, $(3d^{2}+2g^{4})((3d^{2})^{2}-(3d^{2})(2g^{4})+(2g^{4})^{2})=(3d^{2}+2g^{4})(9d^{4}-6d^{2}g^{4}+4g^{8})$.

Answer:

$(3d^{2}+2g^{4})(9d^{4}-6d^{2}g^{4}+4g^{8})$