Question

cube root of 56 x to the fourteenth power.

Answer 1

what are the directions to this? o.o is there any other info to this problem ? what topic / subject does this fall under?

Answer 2

algebra 2; radicals

Answer 3

\[\large (\sqrt[4]{56x})^{14}\]
is this the question?

Answer 4

make that 4 a 3, cubic... yeah

Answer 5

yes

Answer 6

Well, note that this is the same as
\[\large((56x)^{\frac14})^{\frac1{14}}\]
try and remember the rules of exponents, if you get stuck lemme know.

Answer 7

is the answer 2x^4 cube root 7x^2 ?

Answer 8

\[\large((56x)^{\frac13})^{14} \]

Answer 9

thats the answer ?

Answer 10

No, just making sure I have the question right.
I'd break it into
\[ (56x)^{\frac{12}{3}} (56x)^{\frac23} \]

Answer 11

first part gives us (56x)^4

Answer 12

for the second part : (56x)^(2/3)= (7*2^3)^(2/3) = 7^(2/3) * 2^2 = 4* 49^(1/3)

Answer 13

the question is ^3\[\sqrt{56x}\]^14

Answer 14

Yes the problem is
\[(\sqrt[3]{56x})^{14}\]

Answer 15

which is the same as
\[ (56x)^{\frac{14}{3}} \]

Answer 16

\[4(56x)^4\sqrt[3]{49x^2}\]

Answer 17

forgot the x^2

Answer 18

Did I lose you?