help 6^3√(16z^7)+4√(200z^5)

Question

jim_thompson5910 · Answer

Is this 

$$\Large 6^3\sqrt{16z^7}+4\sqrt{200z^5}$$

OR

is it

$$\Large 6\sqrt[3]{16z^7}+4\sqrt{200z^5}$$

anonymous · Answer

the second one

jim_thompson5910 · Answer

alright thank you

anonymous · Answer

thankyou

jim_thompson5910 · Answer

$$\Large 6\sqrt[3]{16z^7}+4\sqrt{200z^5}$$


$$\Large 6\sqrt[3]{8*2z^6*z}+4\sqrt{100*2z^4*z}$$


$$\Large 6\sqrt[3]{8z^6*2z}+4\sqrt{100z^4*2z}$$


$$\Large 6\sqrt[3]{2^3(z^2)^3*2z}+4\sqrt{10^2(z^2)^2*2z}$$


$$\Large 6\sqrt[3]{(2z^2)^3*2z}+4\sqrt{(10z^2)^2*2z}$$


$$\Large 6\sqrt[3]{(2z^2)^3}*\sqrt[3]{2z}+4\sqrt{(10z^2)^2}*\sqrt{2z}$$


$$\Large 6(2z^2)*\sqrt[3]{2z}+4(10z^2)*\sqrt{2z}$$


$$\Large 12z^2\sqrt[3]{2z}+40z^2\sqrt{2z}$$


-----------------------------------------------

So 

$$\Large 6\sqrt[3]{16z^7}+4\sqrt{200z^5}$$

simplifies to 

$$\Large 12z^2\sqrt[3]{2z}+40z^2\sqrt{2z}$$

jim_thompson5910 · Answer

unfortunately we can't go any further because we can't combine the cube and square roots, so we stop here