find the value of
64 to power -1/3 (64 to power 1/3 - 64 power 2/3
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Answered by
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Thanks for kind and nice question,
I hope my explanation is quite understandable,
:)
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Iff (i.e If and only if ) you found it helpful for you.....
Yo ///
:)
I hope my explanation is quite understandable,
:)
And put the batch of glory "Brainliest",
Iff (i.e If and only if ) you found it helpful for you.....
Yo ///
:)
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oceanphotocollection:
thanks
Its k
:)
Humanity :)
pls mark it brainliest if u think it is correct
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Answered by
4
Hi friend!
Here is your Answer!
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![{64}^{ \frac{ - 1}{3} } (64 {}^{ \frac{1}{3} } - 64 {}^{ \frac{2}{3} } ) \\ now \: inverse \: valu \: e \: go \: to \: bottom \: side \\ \\ = \frac{\sqrt[3]{64} - {4}^{3 \times \frac{2}{3} } }{ \sqrt[3]{64} } \\ \\ we \: know \: that \\ \sqrt[3]{64} = 4 \\ then \\ \\ = \frac{4 - {4}^{2} }{4} \\ = \frac{4 - 16}{4} \\ \\ = \frac{ - 12}{4} \\ \\ = - 3 \: ans \\ hope \: its \: helpful](https://tex.z-dn.net/?f=+%7B64%7D%5E%7B+%5Cfrac%7B+-+1%7D%7B3%7D+%7D+%2864+%7B%7D%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D++-+64+%7B%7D%5E%7B+%5Cfrac%7B2%7D%7B3%7D+%7D+%29+%5C%5C+now+%5C%3A+inverse+%5C%3A+valu+%5C%3A+e+%5C%3A+go+%5C%3A+to+%5C%3A+bottom+%5C%3A+side+%5C%5C++%5C%5C++%3D+++%5Cfrac%7B%5Csqrt%5B3%5D%7B64%7D++-++%7B4%7D%5E%7B3+%5Ctimes++%5Cfrac%7B2%7D%7B3%7D+%7D+%7D%7B+%5Csqrt%5B3%5D%7B64%7D+%7D++%5C%5C++%5C%5C+we+%5C%3A+know+%5C%3A+that+%5C%5C++%5Csqrt%5B3%5D%7B64%7D++%3D+4+%5C%5C+then+%5C%5C++%5C%5C++%3D++%5Cfrac%7B4+-++%7B4%7D%5E%7B2%7D+%7D%7B4%7D++%5C%5C++%3D++%5Cfrac%7B4+-+16%7D%7B4%7D++%5C%5C+%5C%5C+++%3D+++%5Cfrac%7B+-+12%7D%7B4%7D++%5C%5C++%5C%5C++%3D++-+3+%5C%3A+ans+%5C%5C+hope+%5C%3A+its+%5C%3A+helpful "{64}^{ \frac{ - 1}{3} } (64 {}^{ \frac{1}{3} } - 64 {}^{ \frac{2}{3} } ) \\ now \: inverse \: valu \: e \: go \: to \: bottom \: side \\ \\ = \frac{\sqrt[3]{64} - {4}^{3 \times \frac{2}{3} } }{ \sqrt[3]{64} } \\ \\ we \: know \: that \\ \sqrt[3]{64} = 4 \\ then \\ \\ = \frac{4 - {4}^{2} }{4} \\ = \frac{4 - 16}{4} \\ \\ = \frac{ - 12}{4} \\ \\ = - 3 \: ans \\ hope \: its \: helpful")
Here is your Answer!
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