# How do you use the Binomial Theorem to expand (3x^2 - (5/x))^3 ?

Sep 5, 2016

${\left(3 {x}^{2} - \frac{5}{x}\right)}^{3} = 27 {x}^{6} - 135 {x}^{3} + 225 - \frac{125}{x} ^ 3$.

#### Explanation:

${\left(a + b\right)}^{n} = {\text{_nC_0*a^(n-0)*b^0+""_nC_1*a^(n-1)*b^1+""_nC_2*a^(n-2)*b^2+""_nC_3*a^(n-3)*b^3+...+}}_{n} {C}_{n} \cdot {a}^{n - n} \cdot {b}^{n}$.

$\therefore {\left(a - b\right)}^{3} = {\text{_3C_0*a^(3-0)*(-b)^0+""_3C_1*a^(3-1)*(-b)^1+""_3C_2*a^(3-2)*(-b)^2+}}_{3} {C}_{3} \cdot {a}^{3 - 3} \cdot {\left(- b\right)}^{3}$.

Knowing that, ${\text{_3C_0=""_3C_3=1, and, ""_3C_1=}}_{3} {C}_{2} = 3$,

${\left(a - b\right)}^{3} = {a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} - {b}^{3.} \ldots \ldots \ldots \ldots \ldots \ldots . . \left(\star\right)$

Letting, $a = 3 {x}^{2} , \mathmr{and} , b = \frac{5}{x} \text{ in } \left(\star\right)$, we get,

${\left(3 {x}^{2} - \frac{5}{x}\right)}^{3} = 27 {x}^{6} - 3 \left(9 {x}^{4}\right) \left(\frac{5}{x}\right) + 3 \left(3 {x}^{2}\right) {\left(\frac{5}{x}\right)}^{2} - \frac{125}{x} ^ 3$

$= 27 {x}^{6} - 135 {x}^{3} + 225 - \frac{125}{x} ^ 3$.

Enjoy maths.!