## anonymous one year ago Evaluate. ^3 sqrt 343 + 3/4 ^3 sqrt -8

1. anonymous

"sqrt" is square root btw

2. anonymous

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

3. anonymous

yes exactly

4. anonymous

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

5. anonymous

how would i plug that in?

6. anonymous

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

7. anonymous

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

8. SolomonZelman

$$\sqrt[3]{343}+\dfrac{3}{4}\sqrt[3]{-8\color{white}{\large|}}$$ 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). $$\sqrt[3]{343}+\dfrac{3}{4}\sqrt[\cancel3]{(-2)^{\cancel 3}\color{white}{\large|} }$$ $$\sqrt[3]{343}+\dfrac{3}{4}(-2)$$

9. SolomonZelman

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.

10. anonymous

i got 5 3/4 but, a. -5 1/2 b. 5 1/2 c. 8 1/2 @SolomonZelman

11. SolomonZelman

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

12. anonymous

7^3 @SolomonZelman