1. **State the problem:** We want to find the limit as $x$ approaches 1 of the expression $$\sqrt[3]{\frac{(3x^2 + 3x - 1)^2}{5x^3 (x^2 - 1)}}.$$\n\n2. **Rewrite the expression:** The expression inside the cube root is a fraction with numerator $(3x^2 + 3x - 1)^2$ and denominator $5x^3 (x^2 - 1)$.\n\n3. **Evaluate numerator and denominator at $x=1$ to check for indeterminate form:**\nNumerator at $x=1$: $(3(1)^2 + 3(1) - 1)^2 = (3 + 3 - 1)^2 = (5)^2 = 25$.\nDenominator at $x=1$: $5(1)^3 (1^2 - 1) = 5(1)(0) = 0$.\n\nSince denominator is zero and numerator is nonzero, the expression tends to infinity or does not exist. But we must check the behavior near $x=1$ carefully.\n\n4. **Factor denominator term $x^2 - 1$:**\n$$x^2 - 1 = (x - 1)(x + 1).$$\n\n5. **Check numerator at $x=1$ more closely:**\n$$3x^2 + 3x - 1 = 3(1)^2 + 3(1) - 1 = 5,$$ so numerator is $25$ at $x=1$.\n\n6. **Since denominator has a factor $(x-1)$ which goes to zero, the fraction inside the cube root tends to infinity or negative infinity depending on the sign near $x=1$.\n\n7. **Check the sign of denominator near $x=1$:**\nFor $x \to 1^+$, $(x-1) > 0$, $x+1 > 0$, $5x^3 > 0$, so denominator $> 0$.\nFor $x \to 1^-$, $(x-1) < 0$, so denominator $< 0$.\n\n8. **Therefore, inside the cube root, the fraction tends to $+\infty$ as $x \to 1^+$ and $-\infty$ as $x \to 1^-$.\n\n9. **Cube root of $+\infty$ is $+\infty$, cube root of $-\infty$ is $-\infty$.\n\n10. **Conclusion:** The limit does not exist because the left and right limits are different (one tends to $+\infty$, the other to $-\infty$).