Question

Evaluate.
^3 sqrt 343 + 3/4 ^3 sqrt -8

Answer 1

"sqrt" is square root btw

Answer 2

Does it look like this \[\bf \sqrt[3]{343}~+~\frac{3}{4} \sqrt[3]{-8}\] ?

Answer 3

yes exactly

Answer 4

Well here is the exponent rule you can use \[\sf \large \sqrt{b^{n}}~=~b^{\frac{1}{n}}\]

Answer 5

how would i plug that in?

Answer 6

Oh whoops I made a mistake it should be \[\sf \large \sqrt[n]{b}~=~b^{\frac{1}{n}}\]

Answer 7

what do i do with the 3/4 just add it at the end?

Answer 8

\(\sqrt[3]{343}+\dfrac{3}{4}\sqrt[3]{-8\color{white}{\large|}}\)

you know that (-8)=(-2)³

\(\sqrt[3]{343}+\dfrac{3}{4}\sqrt[3]{(-2)^3\color{white}{\large|} }\)

you can cancel the powers in the second term and this will give you a -2.
Note: You can do this- to cancel the powers - only when they are odd, because if they are even you would get an absolute value of -2 and might make a mistake (since with even powers that is equivalent to 2, not -2).

\(\sqrt[3]{343}+\dfrac{3}{4}\sqrt[\cancel3]{(-2)^{\cancel 3}\color{white}{\large|} }\)

\(\sqrt[3]{343}+\dfrac{3}{4}(-2) \)

Answer 9

then, 343 = 7³
so you can re-write the 343, and then cancel the powers just the same way I have done it with the second term of your expression.

Answer 10

i got 5 3/4 but,

a. -5 1/2
b. 5 1/2
c. 8 1/2
@SolomonZelman

Answer 11

what did you get for the \(\sqrt[3]{343}\) part, what was this part equivalent to?

Answer 12

7^3 @SolomonZelman