anonymous
  • anonymous
Evaluate. ^3 sqrt 343 + 3/4 ^3 sqrt -8
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
"sqrt" is square root btw
anonymous
  • anonymous
Does it look like this \[\bf \sqrt[3]{343}~+~\frac{3}{4} \sqrt[3]{-8}\] ?
anonymous
  • anonymous
yes exactly

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anonymous
  • anonymous
Well here is the exponent rule you can use \[\sf \large \sqrt{b^{n}}~=~b^{\frac{1}{n}}\]
anonymous
  • anonymous
how would i plug that in?
anonymous
  • anonymous
Oh whoops I made a mistake it should be \[\sf \large \sqrt[n]{b}~=~b^{\frac{1}{n}}\]
anonymous
  • anonymous
what do i do with the 3/4 just add it at the end?
SolomonZelman
  • 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). \(\sqrt[3]{343}+\dfrac{3}{4}\sqrt[\cancel3]{(-2)^{\cancel 3}\color{white}{\large|} }\) \(\sqrt[3]{343}+\dfrac{3}{4}(-2) \)
SolomonZelman
  • 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.
anonymous
  • anonymous
i got 5 3/4 but, a. -5 1/2 b. 5 1/2 c. 8 1/2 @SolomonZelman
SolomonZelman
  • SolomonZelman
what did you get for the \(\sqrt[3]{343}\) part, what was this part equivalent to?
anonymous
  • anonymous
7^3 @SolomonZelman

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