The term independent of x in the expansion of (√[6]x-(1/√[3]x))9 is

Question

The term independent of x in the expansion of (√[6]x-(1/√[3]x))9 is (A) - 9 C 3 (B) - 9 C 4 (C) - 9 C 5 (D) - 8 C 3

Tardigrade Admin · Accepted Answer

T r +1= 9 C r (√[6] x )9- r (-(1/√[3] x )) r = 9 C r (-1) r . x (9- r /6)-( r /3)= 9 C r .x ((9-3 r /6)) Now (9-3 r/6)=0 ⇒ r=3 Thus, term independent of x =- 9 C 3

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

$-^{9} C_{3}$

$-^{9} C_{4}$

$-^{9} C_{5}$

$-^{8} C_{3}$

$T_{r + 1} =^{9} C_{r} (6 x)^{9 - r} (- \frac{1}{3 x})^{r}$
$=^{9} C_{r} (- 1)^{r} . x^{\frac{9 - r}{6} - \frac{r}{3}} =^{9} C_{r} . x^{(\frac{9 - 3 r}{6})}$
Now $\frac{9 - 3 r}{6} = 0$
$\Rightarrow r = 3$
Thus, term independent of $x = -^{9} C_{3}$

Q. The term independent of x in the expansion of (6x​−3x​1​)9 is

−9C3​

−9C4​

−9C5​

−8C3​