Simplify $2 \sqrt[3]{50x^2 z^5} × 3 \sqrt[3]{15 y^3 z}$ Image of problem
As you can see the answer is $30 y z^2 \sqrt[3]{6x^2}$ . I understand mostly everything in the problem, but one thing that I am having confusion on is where the "30" came from in the answer. I tried many ways to find out how the "30" got there.
 A: You have:
$$6 \sqrt[3]{5^2\cdot 2 \cdot x^2 \cdot z^5 \cdot 3 \cdot 5 \cdot y^3 \cdot z} = 6 \sqrt[3]{5^3\cdot 6 \cdot x^2 \cdot z^6  \cdot y^3} = 30 \cdot y \cdot z^2  \sqrt[3]{ 6 \cdot x^2} $$
I hope its clear now.
A: $2 \sqrt[3]{50x^2 z^5} * 3 \sqrt[3]{15 y^3 z}$
The first thing I recommend doing is to notice that the radicals are of the same power.  This means I should take the coefficients of the radicals and multiply them, then take the things under the radical and multiply them.
$(2*3) \sqrt[3]{(50x^2z^5)*(15y^3z)}=6\sqrt[3]{750x^2y^3z^6}$
Now simplify by factoring out everything you can from under the radical.
$6\sqrt[3]{750x^2y^3z^6}=6\sqrt[3]{(125*6)x^2y^3z^6}==6\sqrt[3]{125y^3z^6}\sqrt[3]{6x^2}=6*5yz^2\sqrt[3]{6x^2}=30yz^2\sqrt[3]{6x^2}=$
The 30 came from a 6*5.  The 6 came from the coefficients of the radical and the 5 came from a factoring of 125 out of 750, which was cube rooted into 5.
A: HINT: we have $$6\cdot \sqrt[3]{750x^2y^3z^6}=30\cdot 3^{1/3}2^{1/3}y\cdot z^{4/3}(x^2z^2)^{1/3}$$
A: $$2\,\sqrt[3]{50\,x^2 z^5} \cdot 3\,\sqrt[3]{15\,\color{red}{y^3} z}=$$
$$=2 \cdot 3 \cdot \color{red}{y} \cdot \sqrt[3]{50\,x^2 z^5} \cdot \sqrt[3]{15\,z}=$$
$$=6\,y \cdot \sqrt[3]{(50\,x^2 z^5)(15 z)}=$$
$$=6\,y \cdot \sqrt[3]{750\,x^2 \color{magenta}{z^6}}=$$
$$=6\,y\color{magenta}{z^2} \cdot \sqrt[3]{2 \cdot 3 \cdot \color{blue}{5^3} \cdot x^2}=$$
$$=6 \cdot \color{blue}{5} \cdot yz^2 \cdot \sqrt[3]{2 \cdot 3 \cdot x^2}=$$
$$=30\,yz^2 \sqrt[3]{6\,x^2}.$$
As you see, $30$ is due to $6\cdot\sqrt[3]{5^3}$, emphatized in blue.
