what is the factored form of $125a^{6}-64$?\n$(25a^{2}+16)(25a^{4}+20a^{2}-4)$\n$(5a^{2}-4)(25a^{4}+20a^{2}+1…

what is the factored form of $125a^{6}-64$?\n$(25a^{2}+16)(25a^{4}+20a^{2}-4)$\n$(5a^{2}-4)(25a^{4}+20a^{2}+16)$\n$(25a^{2}-16)(25a^{4}-20a^{2}+16)$\n$(5a^{2}-4)(25a^{4}-20a^{2}+16)$

what is the factored form of $125a^{6}-64$?\n$(25a^{2}+16)(25a^{4}+20a^{2}-4)$\n$(5a^{2}-4)(25a^{4}+20a^{2}+16)$\n$(25a^{2}-16)(25a^{4}-20a^{2}+16)$\n$(5a^{2}-4)(25a^{4}-20a^{2}+16)$

Answer

Explanation:

Step1: Recognize difference - of - cubes

The expression $125a^{6}-64$ is in the form of $x^{3}-y^{3}$, where $x = 5a^{2}$ ($(5a^{2})^{3}=125a^{6}$) and $y = 4$ ($4^{3}=64$). The formula for factoring $x^{3}-y^{3}$ is $(x - y)(x^{2}+xy + y^{2})$.

Step2: Substitute $x$ and $y$

Substitute $x = 5a^{2}$ and $y = 4$ into the formula. $x - y=5a^{2}-4$. $x^{2}=(5a^{2})^{2}=25a^{4}$, $xy=(5a^{2})\times4 = 20a^{2}$, $y^{2}=4^{2}=16$. So $x^{2}+xy + y^{2}=25a^{4}+20a^{2}+16$. The factored form of $125a^{6}-64$ is $(5a^{2}-4)(25a^{4}+20a^{2}+16)$.

Answer:

$(5a^{2}-4)(25a^{4}+20a^{2}+16)$