So we have:
[tex]-3a^5b^2\sqrt[3]{a^2c}[/tex]And we want to knowthe original before simplification, that is, before evaluating the interior part of the root.
So, we need to figure a way to put the part outside of the root back in.
Taking the cubic root of a number is the same as dividing its exponent by 3, because:
[tex]\sqrt[3]{a^n}=a^{\frac{n}{3}}[/tex]So, thinking in the other direction, we need to multiply the exponents by 3 before taking it back to the inside of the cubic root:
[tex]a^k=a^{\frac{3k}{3}}=\sqrt[3]{a^{3k}}[/tex]So, the b part have a 2 in the exponent, so we can multiply it by 3 to get 6:
[tex]\begin{gathered} b^2=b^{\frac{3\cdot2}{3}}=\sqrt[3]{b^{3\cdot2}}=\sqrt[3]{b^6} \\ -3a^5b^2\sqrt[3]{a^2c}=-3a^5\sqrt[3]{b^6}\sqrt[3]{a^2b^{3\cdot2}c}=-3a^5\sqrt[3]{a^2b^6c} \end{gathered}[/tex]The a part have a 5 in the exponent, so we will get 15:
[tex]\begin{gathered} a^5=a^{\frac{3\cdot5}{3}}=\sqrt[3]{a^{3\cdot5}}=\sqrt[3]{a^{15}} \\ -3a^5\sqrt[3]{a^2b^6c}=-3\sqrt[3]{a^{15}}\sqrt[3]{a^2^{}b^6c}=-3\sqrt[3]{a^2a^{15}b^6c} \end{gathered}[/tex]Now, since we have a² and a¹⁵, we can add their exponents:
[tex]\begin{gathered} a^2a^{15}=a^{17} \\ -3\sqrt[3]{a^2a^{15}b^6c}=-3^{}\sqrt[3]{a^{17}b^6c} \end{gathered}[/tex]Now, the -3 have an exponent of 1, so:
[tex]\begin{gathered} -3=(-3)^1=(-3)^{\frac{3\cdot1}{3}}=\sqrt[3]{(-3)^{3\cdot1}}=\sqrt[3]{(-3)^3}=\sqrt[3]{-27} \\ -3^{}\sqrt[3]{a^{17}b^6c}=\sqrt[3]{-27}^{}\sqrt[3]{a^{17}b^6c}=^{}\sqrt[3]{-27a^{17}b^6c} \end{gathered}[/tex]Thus, we have, in the end:
[tex]^{}\sqrt[3]{-27a^{17}b^6c}=-3a^5b^2^{}\sqrt[3]{a^2^{}c}[/tex]