# How do you use the binomial formula to find the coefficient of x^3 in (3x^2 - (5/x))^3?

Sep 28, 2016

$\text{The reqd. co-eff.=} - 135.$

#### Explanation:

In the expansion of ${\left(a + b\right)}^{n}$, the ${\left(r + 1\right)}^{t h}$ term, i.e., ${T}_{r + 1}$ is,

T_(r+1)=""_nC_ra^(n-r)b^r.

In our Problem, we have, $a = 3 {x}^{2} , b = - \frac{5}{x} , \mathmr{and} , n = 3.$ So,

T_(r+1)=""_3C_r(3x^2)^(3-r)(-5/x)^r.

=(-5)^r""_3C_r(3)^(3-r)x^(6-2r)x^-r.

=(-5)^r(3)^(3-r)""_3C_rx^(6-3r)

Hence, if ${T}_{r + 1}$ is the term containing ${x}^{3}$, we must have,

$6 - 3 r = 3 \Rightarrow r = 1$

Therefore, the reqd. co-eff. is

(-5)^1(3)^(3-1)""_3C_1=-5*9*3=-135.

Enjoy Maths.!