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How can we find the term independent of $x$ in the expansion of,

$(x^{\frac{2}{3}}+4x^{\frac{1}{3}}+4)^5\cdot\left(\dfrac{1}{x^{\frac{1}{3}}-1}+\dfrac{1}{x^{\frac{2}{3}}+x^{\frac{1}{3}}+1}\right)^{-9}$

Thank you in advance !

2 Answers2

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Hint: The expression is equal to (use some algebraic identities): $$\frac{(x^{1/3}+2)(x-1)^9}{x^3}=2\binom93(-1)^6+\text{other terms with non-zero powers of x}$$

Further Hint: $$(x-1)(x^2+x+1)=(x^3-1)$$

RE60K
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  • Yes, the same at the same time !! When I see monsters, I know (look at my age !) that there must be a simple trick. Happy New Year ! – Claude Leibovici Dec 31 '14 at 13:29
  • @AlKhwarizmi. ADG and I gave the same answer ! – Claude Leibovici Dec 31 '14 at 13:32
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    Well... I was not being able to simplify the equation to $\dfrac{\left(\sqrt[3]{x}+2\right) (x-1)^9}{x^3}$ –  Dec 31 '14 at 13:33
  • Okay, Now I am getting the constant term $9\choose9$$+$$9\choose3$$(-1)^6$, please help –  Dec 31 '14 at 13:47
  • we need to find constant term in $\frac{(x^{1/3}+2)(x-1)^9}{x^3}$ or the coefficient of $x^3$ in $(x^{1/3}+2)(x-1)^9$; since the first bracket doesn't give any $x^3$ but only a non- integer term which when combined with a integral term term of second would not give $x^3$ so only choice remains $2$ from bracket 1 and coefficient of x^3 from bracket 2, and as the termof $x^k$ from second bracket is $\binom 9kx^k(-1)^{9-k}$, we're done. @AlKhwarizmi – RE60K Dec 31 '14 at 13:53
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Hint

$$\dfrac{1}{x^{\frac{1}{3}}-1}+\dfrac{1}{x^{\frac{2}{3}}+x^{\frac{1}{3}}+1}=\frac{\left(\sqrt[3]{x}+2\right) \sqrt[3]{x}}{x-1}$$ $$x^{\frac{2}{3}}+4x^{\frac{1}{3}}+4=\left(\sqrt[3]{x}+2\right)^2$$ $$(x^{\frac{2}{3}}+4x^{\frac{1}{3}}+4)^5\cdot\left(\dfrac{1}{x^{\frac{1}{3}}-1}+\dfrac{1}{x^{\frac{2}{3}}+x^{\frac{1}{3}}+1}\right)^{-9}=\frac{\left(\sqrt[3]{x}+2\right) (x-1)^9}{x^3}$$

I am sure that you can take from here.

Happy New Year

Claude Leibovici
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