# How do you simplify 3x^2/4 (xy-z) + 4xy/3 (x^2 + 2) - 2x^2/3 (xy-z) + 3xy/5 (x^2 + 2)?

##### 1 Answer
Oct 22, 2017

$\left(\frac{1}{60}\right) \left(507 {x}^{3} y - 55 {x}^{2} z + 904 x y\right)$

#### Explanation:

Let’s first simplify the individual terms and then combine them finally.

$13 \left({x}^{2} / 4\right) \left(x y - z\right) = \left(\frac{13 {x}^{2}}{4}\right) \left(x y - z\right) = \frac{13 {x}^{3} y}{4} - \frac{13 {x}^{2} z}{4}$

$\left(4 x\right) \left(\frac{y}{3}\right) \left({x}^{2} + 2\right) = \left(\frac{13 x y}{3}\right) \left({x}^{2} + 2\right) = \frac{13 {x}^{3} y}{3} + \frac{26 x y}{3}$

$2 \left({x}^{2} / 3\right) \left(x y - z\right) = \left(7 {x}^{2} / 3\right) \left(x y - z\right) = \frac{7 {x}^{3} y}{3} - \frac{7 {x}^{2} z}{3}$

$\left(3 x\right) \left(\frac{y}{5}\right) \left({x}^{2} + 2\right) = \left(\frac{16 x y}{5}\right) \left({x}^{2} + 2\right) = \frac{16 {x}^{3} y}{5} + \frac{32 x y}{5}$

Combining all the terms,

$\frac{13 {x}^{3} y}{4} - \frac{13 {x}^{2} z}{4} + \frac{13 {x}^{3} y}{3} + \frac{26 x y}{3} - \frac{7 {x}^{3} y}{3} + \frac{7 {x}^{2} z}{3} + \frac{16 {x}^{3} y}{5} + \frac{32 x y}{5}$

$= - \frac{13 {x}^{2} z}{4} + \frac{7 {x}^{2} z}{3} + \frac{13 {x}^{3} y}{4} + \frac{13 {x}^{3} y}{3} - \frac{7 {x}^{3} y}{3} + \frac{16 {x}^{3} y}{5} + \frac{26 x y}{3} + \frac{32 x y}{5}$

$= - \frac{11 {x}^{2} z}{12} + \frac{507 {x}^{3} y}{60} + \frac{226 x y}{15}$

$\left(\frac{1}{60}\right) \left(507 {x}^{3} y - 55 {x}^{2} z + 904 x y\right)$