Question

Can you help solve this by simplifying the radical?

Answer 1

First of all, we can split the root of the multiplication into the multiplication of the roots:

$\sqrt[3]{-56ab^6c^{10}}=\sqrt[3]{-56}\cdot\sqrt[3]{a}\cdot\sqrt[3]{b^6}\cdot\sqrt[3]{c^10}$

If we look at the prime factorization of -56 we have

$-56=2^3\cdot 7$

So, we have

$\sqrt[3]{-56} = -\sqrt[3]{2^3\cdot 7} = -\sqrt[3]{2^3}\cdot\sqrt[3]{7} = -2\sqrt[3]{7}$

We can do nothing about $\sqrt[3]{a}$, because the exponent is lower than the order of the root.

We have

$\sqrt[3]{b^6} = (b^6)^{\frac{1}{3}} = b^\frac{6}{3}=b^2$

Finally, we have

$\sqrt[3]{c^{10}} =\sqrt[3]{c^{9+1}}=\sqrt[3]{c^{9}\cdot c}=c^\frac{9}{3}\cdot\sqrt[3]{c}=c^3\sqrt[3]{c}$

So, the whole expression is equivalent to

$-2b^2c^3\sqrt[3]{7ac}$