Question

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

`(6x^5y^2-16x^4y^3-6x^3y^4-9x^2+24xy+9y^2)/(18y^2)`

Explanation:

The applicable rule of exponents is ...

(a^b)(a^c) = a^(b+c)

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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.

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