what is the factored form of $8x^{24}-27y^{6}$?\n$(8x^{8}-27y^{2})(2x^{16}+xy + 3y^{4})$\n$(2x^{8}-3y^{2})(4x…

what is the factored form of $8x^{24}-27y^{6}$?\n$(8x^{8}-27y^{2})(2x^{16}+xy + 3y^{4})$\n$(2x^{8}-3y^{2})(4x^{16}-6x^{8}y^{2}+9y^{4})$\n$(2x^{8}-3y^{2})(4x^{16}+6x^{8}y^{2}+9y^{4})$\n$(8x^{8}-27y^{2})(2x^{16}-6xy + 3y^{4})

what is the factored form of $8x^{24}-27y^{6}$?\n$(8x^{8}-27y^{2})(2x^{16}+xy + 3y^{4})$\n$(2x^{8}-3y^{2})(4x^{16}-6x^{8}y^{2}+9y^{4})$\n$(2x^{8}-3y^{2})(4x^{16}+6x^{8}y^{2}+9y^{4})$\n$(8x^{8}-27y^{2})(2x^{16}-6xy + 3y^{4})

Answer

Explanation:

Step1: Recognize difference - of - cubes formula

The expression $8x^{24}-27y^{6}$ is in the form of $a^{3}-b^{3}$, where $a = 2x^{8}$ ($(2x^{8})^{3}=8x^{24}$) and $b = 3y^{2}$ ($(3y^{2})^{3}=27y^{6}$). The difference - of - cubes formula is $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$.

Step2: Substitute values into formula

Substitute $a = 2x^{8}$ and $b = 3y^{2}$ into the formula. We get $(2x^{8}-3y^{2})((2x^{8})^{2}+(2x^{8})(3y^{2})+(3y^{2})^{2})$.

Step3: Simplify the second factor

$(2x^{8})^{2}=4x^{16}$, $(2x^{8})(3y^{2}) = 6x^{8}y^{2}$, and $(3y^{2})^{2}=9y^{4}$. So the factored form is $(2x^{8}-3y^{2})(4x^{16}+6x^{8}y^{2}+9y^{4})$.

Answer:

$(2x^{8}-3y^{2})(4x^{16}+6x^{8}y^{2}+9y^{4})$