Expand each of the expressions.
$$
\left(\frac{x}{3}+\frac{1}{x}\right)^{5}
$$

Answer

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Algebra

NCERT Class 11 - Math

Chapter 8

Binomial Theorem

Section 1

Introduction

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Video Transcript

So in the during question, we are told to find the expansion of X by three plus one by X raised to the power fight. And now we are told to use the binomial theorem. So, according to the binomial theorem, when we take the expansion of A plus B raised to the power end, what we would have is and C zero times eight to the power and less. And see one times A to the power and minus one times deep plus n C two eight to the power and minus to be square plus, etcetera. We have N C n minus one times A times to be to the poet and minus one. Press N C N plus N C N times lead to the bar. And so this is the expansion that we have that regard using the binomial terram. And we're going to use this formula which as a result of binomial theorem, in order to find the expansion of the given expression, right? So what we would have over here is N. S equal to five A. And B. R. A. Is equal to its by three and B is equal to one by X. So what we would have is fine. C zero X by three raised to the power five plus five C one X might be raised to the power forward times one by X plus five C. Two X by three. Raised to the power three times one by X. Raised to the power to plus five C. Three X by three. Raised to the power to one by X. Raised to the power three. Plus we have five C. Four X by two. Raised to the power 11 by X. Raised to the powerful plus we have the final damages. Five C five X by two raised to the power zero which is equal to one times one by X. Raised to the power of five. And now they can't right this expansion as 50 is equal to one. So it would have the same value of five C five. And next we have five C. One which is equal to five which would also be the value of five C. Four. Then we have five C two which is equal to 10 which is also the valuable five C. Three. So having these values in mind now we can sub stored them in the expansion and what we would have is fine. C zero is one. So the first term is X to the power five divided by three to the power five which is equal to three to the power five years 243 plus we have five times X to the power forward divided by X which is five X cube divided by three to the powerful which is equal to 81 plus we have five C two which is equal to 10 10 times X cubed divided by X square as X divided by three cube is equal to three cube. S 27. The next term S five C three times five C three is 10 10 times X squared divided by x cube is 10 by X. And we have in the denominator we have three square which is equal to mine. So what we haven't stand by nine X as the uh fourth 12 3rd time. Right next we have the fourth tone which is five C four which is equal to 55 times X divided by extremely powerful is equal to X squared divided by X cubed. Basic will to one by X. X. Raised to the power it one divided by X. To the power forward is one by X cube. So what we would have us five divided by two to the power for 16 five divided by 16 X. Cube. And the last time as five C. Five which is equal to one. And what we would have over here is one to the power one by extra power five. So this is what we have as the expansion of the given expression will be found using the binomial Tierro. Right? So I hope you understand the method. Thank you. Oh.

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