Question

please solve this question​

Answer 1

Answer:

$(\frac{-y^9*x }{18} )$

Step-by-step explanation:

Find 4th term in the expansion of (3x - $\frac{y^3} {6}$ )⁴

General term expansion of (a+b)ⁿ is given by the formula,

$T_{r+1}$ = ⁿC$_{r}$ $a^{n-r}$ $b^{r}$

=> $T_{4}$ = $T_{3+1}$ = ⁴C$_{3}$ * $(3x)^{4-3}$* $(\frac{-y^3}{6} )^{3}$

                    = $\frac{4! }{3! (4-3)!}$ * $(3x)^{1}$* $(\frac{-y^9}{6^3} )$

                    = 4 * $(3x)^{1}$* $(\frac{-y^9}{216} )$

                    = x * $(\frac{-y^9}{18} )$

                    = $(\frac{-y^9*x }{18} )$