Question

the product of (3/2xyz - 9/4xy^2z^3) and (-8/27xyz)​

Answer 1

Answer:

$2x^{2} y^{2} z^{2} (\frac{-2+3yz^{2} }{9})$

Step-by-step explanation:

The equation understood by me:

$(\frac{3xyz}{2} -\frac{9xy^{2}z^{3}  }{4})*(\frac{-8xyz}{27}  )$

So, if I got the equation correct then the solution is right......

1. Remove the Brackets or Split the Brackets :

$(\frac{3xyz}{2}*\frac{-8xyz}{27} )-(\frac{9xy^{2}z^{3} }{4}*\frac{-8xyz}{27} )$

2. Multiply the Brackets further :

$(\frac{-4x^{2} y^{2} z^{2} }{9} )-(\frac{-2x^{2} y^{3} z^{4} }{3})$

3. Remove the negative sign in middle of the two terms (Remove the negative sign of the numerator of the second term and the negative sign in middle of the two terms and replace with positive sign).

4. Take LCM of the Denominators :

$\frac{-4x^{2} y^{2} z^{2} + 6x^{2} y^{3} z^{4} }{9}$

5. Take 2$x^{2}y^{2} z^{2}$ as common outside the Bracket :

$2x^{2} y^{2} z^{2} (\frac{-2+3yz^{2} }{9})$

There you go!! The Answer is Step 5......