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2 cubed root 81+3 cubed root 24, 2sqrt[4]+3cubed root(8)(3)
- Thread starter Al royal
- Start date
D
Deleted member 4993
Guest
Re: hey guys
Are you saying
\(\displaystyle {81}^{\frac{1}{3}} = \sqrt{4}\) .........That is definitely incorrect
What are the prime factors of 81?
Al royal said:I ran into a little problem with a few questions.
2 cubed root 81 + 3 cubed root 24
2?4 + 3 cubed root?(8)(3)
have no idea where to go from here....
thanks in advanced
-AL
Are you saying
\(\displaystyle {81}^{\frac{1}{3}} = \sqrt{4}\) .........That is definitely incorrect
What are the prime factors of 81?
Re: hey guys
I ran into a little problem with a few questions.
2 cubed root 81 + 3 cubed root 24
2?4 + 3 cubed root?(8)(3)
have no idea where to go from here....
thanks in advanced
I guess you mean...
\(\displaystyle 2\sqrt[3]{81} + 3\sqrt[3]{24}\)
\(\displaystyle 2\sqrt{4} + 3\sqrt[3]{(8)(3)}\)
Neither do I. Is this one problem or two? What are you trying to do? Are you combining the two expressions? Are you simplifying? Just what is your question and how far have you gotten? Where are you stuck?
I ran into a little problem with a few questions.
2 cubed root 81 + 3 cubed root 24
2?4 + 3 cubed root?(8)(3)
have no idea where to go from here....
thanks in advanced
I guess you mean...
\(\displaystyle 2\sqrt[3]{81} + 3\sqrt[3]{24}\)
\(\displaystyle 2\sqrt{4} + 3\sqrt[3]{(8)(3)}\)
Neither do I. Is this one problem or two? What are you trying to do? Are you combining the two expressions? Are you simplifying? Just what is your question and how far have you gotten? Where are you stuck?
Re: hey guys
thats exactly what i mean..adding both and simplifying.
im not sure on what steps to take next.
Loren said:I ran into a little problem with a few questions.
2 cubed root 81 + 3 cubed root 24
2?4 + 3 cubed root?(8)(3)
have no idea where to go from here....
thanks in advanced
I guess you mean...
\(\displaystyle 2\sqrt[3]{81} + 3\sqrt[3]{24}\)
\(\displaystyle 2\sqrt{4} + 3\sqrt[3]{(8)(3)}\)
Neither do I. Is this one problem or two? What are you trying to do? Are you combining the two expressions? Are you simplifying? Just what is your question and how far have you gotten? Where are you stuck?
thats exactly what i mean..adding both and simplifying.
im not sure on what steps to take next.
D
Deleted member 4993
Guest
First simplify the terms under radical sign. eg
\(\displaystyle \sqrt[3]{8} \, = \, \sqrt[3]{2^3} \, = 2\)
and
\(\displaystyle \sqrt[3]{81} \, = \, ???\)
\(\displaystyle \sqrt{4} \, = \, ???\)
and so on...
\(\displaystyle \sqrt[3]{8} \, = \, \sqrt[3]{2^3} \, = 2\)
and
\(\displaystyle \sqrt[3]{81} \, = \, ???\)
\(\displaystyle \sqrt{4} \, = \, ???\)
and so on...
D
Deleted member 4993
Guest
Al royal said:how exactly did u get 3 cubed root of 8?
do you understand:
what does cube - like 2[sup:r5nbh1h2]3[/sup:r5nbh1h2] - mean?
what does cube-root - like 8[sup:r5nbh1h2](1/3)[/sup:r5nbh1h2] - mean?