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  4. The term independent of y in the expansion of (y 1/6 -y -1/3)9 is

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Q. The term independent of $y$ in the expansion of $ ({y} ^{1/6} -y ^{-1/3})^9$ is

Binomial Theorem

Solution:

$T_{r+1} = \,{}^{9}c_{r} \left(y^{\frac{1}{6}}\right)^{9-r} \left(-y^{\frac{1}{3}}\right)^{r}$
$= \,{}^{9}c_{r}y ^{\frac{9-r}{6}} \cdot \left(-1\right)^{r}\,y^{\frac{-r}{3}}$
$\therefore \frac{9-r}{6} - \frac{r}{3} = 0$
$\Rightarrow r = 3$
$\therefore $ reqd. term $= \,{}^{9}c_{3} \left(- 1\right)^{3} = -\,{}^{9}c_{3}$
$= \frac{9 \times 8 \times 7}{1 \times 2 \times 3}$
$= - 3 \times 4\times 7 = -84$