theeledz
Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Please help simplify.The variables have no values.
(2/3x³-1/y²)(1/2x²-4/3xy-1/2y²)(y≠0)
​

Sagot :

sqdancefan

Answer:

  [tex]\dfrac{6x^5y^2-16x^4y^3-6x^3y^4-9x^2+24xy+9y^2}{18y^2}[/tex]

Step-by-step explanation:

The applicable rule of exponents is ...

  (a^b)(a^c) = a^(b+c)

__

For this problem, it is convenient to factor out a factor that will eliminate fractions. Then the distributive property and the rules for adding fractions apply.

  [tex]\left(\dfrac{2}{3}x^3-\dfrac{1}{y^2}\right)\left(\dfrac{1}{2}x^2-\dfrac{4}{3}xy-\dfrac{1}{2}y^2\right)\qquad\text{given}\\\\=\dfrac{1}{18y^2}(2x^3y^2-3)(3x^2-8xy-3y^2)\\\\=\dfrac{1}{18y^2}(2x^3y^2(3x^2-8xy-3y^2)-3(3x^2-8xy-3y^2))\\\\=\boxed{\dfrac{6x^5y^2-16x^4y^3-6x^3y^4-9x^2+24xy+9y^2}{18y^2}}[/tex]

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.