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If $f(x)=(3x-1)^4(\sqrt[3]{2x-3})^2$, find $f'(-1)$

My attempt,

I differentiated and I got $\frac{2268x^4+3780x^2-5292x^3-1092x+112}{3\sqrt[3]{2x-3}}$, so I substitute $-1$ and I got $\frac{12544}{3\sqrt[3]{-5}}$, but the answer given is $-\frac{12544\sqrt[3]{25}}{15}$. Can anyone explain to me how to simplify into given answer? Thanks in advance.

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    what Kind of function is $$h$$? – 2017-02-02
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    sorry, it was a typo – 2017-02-02
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    It happens that $$\frac{12544}{3\sqrt[3]{-5}}=-\frac{12544\sqrt[3]{25}}{15}$$ if $\sqrt[3]{-5}$ is interpreted as $-\sqrt[3]{5}$. To see why, multiply the numerator and the denominator by $$\sqrt[3]{25}$$ and prove the identity $$3\cdot\sqrt[3]{25}\cdot\sqrt[3]{5}=3\cdot5=15$$ – 2017-02-02

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Rationalize the denominator: multiply and divide by $\sqrt[3]{25}$. Also, when you have "complicated" functions that invoke the use of both the product and chain rules at the same time, it's best to leave everything factored - expanding leads to a mess sometimes.