Question

The term independent of x in the expansion of ...(Q inside)

Answer 1

\[(\frac{ x+1 }{ x^{2/3}-x^{1/3}+1 } - \frac{ x-1 }{ x-x^{1/2} })^{10}\] is ?

Answer 2

Simply write x as t^6 and solve the expression...

Answer 3

\[(\frac{ t^6 +1 }{ t^4-t^2 +1 } - \frac{ t^6 - 1 }{ t^6 - t^3 } )^{10}\]
then?

Answer 4

Apply the identity for a^3+ b^3 and solve the first term

Answer 5

Thr first term will be (1+t)

Answer 6

first term is (1+t^2) right?

Answer 7

@samigupta8

Answer 8

Yep sorry

Answer 9

np ..and for the second term i am getting (t^4+t^2+1 / t^3)

Answer 10

There u have to apply a^2-b^2

Answer 11

ok..i am getting (t^3+1)/t^2.. @samigupta8

Answer 12

I am getting (1+t^-3)

Answer 13

oh i made a mistake..

Answer 14

try to use these two things to simplify your expression-

\((x)^{\frac{1}{3}}+1^3=(x^{\frac{1}{3}}+1)(x^{\frac{2}{3}}-x^{\frac{1}{3}}+1) \)

\(x-1=(\sqrt{x}+1)(\sqrt{x}-1)\)

Answer 15

it should be (x+1) on the left side...

Answer 16

in my 2nd equation?

Answer 17

no first.. anyway its just a typing error no problem..

Answer 18

i am getting:
\[x^{1/3} - (x^{1/2} +1)/\sqrt{x}\]

Answer 19

oh yea got ya :)
\(((x)^{\frac{1}{3}})^{\color{red}3} +1^3=(x^{\frac{1}{3}}+1)(x^{\frac{2}{3}}-x^{\frac{1}{3}}+1)\)

Answer 20

sry.. it is x^1/3 +1

Answer 21

first term in my expression..

Answer 22

yes correct
now do the next substitution and simplify the thing

Answer 23

what substitution ?

Answer 24

i mean
rn we have this-
\(\large (x^{\frac{1}{3}}+1) - \underbrace{\frac{x-1}{x-\sqrt{x}}}\)
now putting \(x-1=(\sqrt{x}+1)(\sqrt{x}-1)\)

we can reduce the whole expression to just 2 terms after the simplification

Answer 25

i have done that .. see abv

Answer 26

sry i didn't read any of that x.x
the problem is already solved?

Answer 27

how? its still lengthy..

Answer 28

well after simplification we have this-

\[\left[ \left( x^\frac{ 1 }{ 3 } +1 \right)-\left( 1+x ^\frac{ -1 }{ 2} \right)\right]^{10}\]
\[\left[ x^\frac{ 1 }{ 3 }-x^{-\frac{1}{2} }\right]^{10}\]

just write the general term and put the power=0

Answer 29

oh ok got it!

Answer 30

Thanks!

Answer 31

np :]

Answer 32

@imqwerty what about the root x in the denominator in the second term..?

Answer 33

i wrote \(\large \frac{\sqrt{x}+1}{\sqrt{x}}\) as \(\large1+ \frac{1}{ \sqrt{x}}\) =\(\large 1+x^\frac{-1}{2}\)

Answer 34

of course!

Answer 35

yes
this kills the denominator :)

Answer 36

yeah now understood...nice idea..