$\sqrt[3]{-125a^6b^{12}}$

Key concept: This is a standard radical simplification problem. The key is to split the number and variables into perfect cubes, then take the cube root of each factor.

Step by step

Step 1: Rewrite the radicand as perfect cubes

= \sqrt[3]{(-5)^3(a^2)^3(b^4)^3}$$ because - $-125 = (-5)^3$ - $a^6 = (a^2)^3$ - $b^{12} = (b^4)^3$ ### Step 2: Take the cube root of each factor $$\sqrt[3]{(-5)^3(a^2)^3(b^4)^3} = -5a^2b^4$$ ### Final answer $$\boxed{-5a^2b^4}$$ ### Pitfall alert A common mistake is to treat cube roots like square roots and forget that odd roots preserve the sign of a negative number. Since this is a cube root, $\sqrt[3]{-125}=-5$, not $+5$. ### Try different conditions If the exponent on $b$ were not a multiple of 3, you would split it into a perfect cube times a leftover factor. For example, $\sqrt[3]{b^{13}}=b^4\sqrt[3]{b}$. ### Further reading cube root, perfect cube, radical simplification