# How do you simplify (8y^3)(-3x^2y^2)(3/8xy^4)?

Jan 10, 2018

$- 9 {x}^{3} {y}^{9}$

#### Explanation:

Well, since multiplication is commutative we can change the order and group similar variables together and constants together, so this is same as
$\left(8\right) \left(- 3\right) \left(\frac{3}{8}\right) \left({y}^{3} {y}^{2} {y}^{4}\right) \left({x}^{2} x\right)$
Then we can cancel $\frac{8}{8}$ since $= 1$ and left with $\left(- 3\right) \left(3\right) = - 9$ for constant part. When multiplying powers with same base we add exponents so $\left({y}^{3} {y}^{2} {y}^{4}\right) = {y}^{9}$
And $\left({x}^{2} x\right) = {x}^{3}$.

Putting it all back together gives $- 9 {x}^{3} {y}^{9}$

Jan 10, 2018

$\textcolor{b l u e}{- 9 {x}^{3} {y}^{9}}$

#### Explanation:

Multiply it out:
$\left(8 {y}^{3}\right) \left(- 3 {x}^{2} {y}^{2}\right) \left(\frac{3}{8} x {y}^{4}\right)$
When multiplying variables, add the exponents:
$\left(- 24 {x}^{2} {y}^{5}\right) \left(\frac{3}{8} x {y}^{4}\right)$
$\textcolor{b l u e}{- 9 {x}^{3} {y}^{9}}$