Verified answer

3√(x^0)(y^4)/3z^3 X 3√(81x^9)(y^-10)(z^6)

x^0 = 1

√y^4 = y^2

√81x^9 = 9x^4√x

√y^(-10) = y^(-5)

= (3 × y^2)/3z^3 X 3 × 9x^4 y^(-5) z^6

= (y^2)/z^3 X 27x^4√x y^(-5) z^6

= (y^2) × 27x^4√x × y^(-5) z^3

= (y^2-5) × 27x^(4+1/2) z^3

= (y^-3) × 27x^(9/2) z^3

= 27√x^(9) z^3/= (y^3)

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It would be easier to read if you used parenthesis.

Is all of x^0y^4/3z^3 and all of 81x^9y^-10z^6 under the root signs. If so write as

3√(x^0y^4/3z^3) * 3√(81x^9y^-10z^6)

Also, using * for multiplication avoids confusion with the variable x

Now I'm not sure if this is 3 TIMES square root (x^0y^4/3z^3) ...

or if 3√ means cube root.

I suspect cube root, because after multiplying everything we end up with all cubes under the root sign.

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Case 1: square root

3√(x⁰y⁴/3z³) * 3√(81x⁹y⁻¹⁰z⁶)

= 3 * 3 * √[(x⁰y⁴/3z³)*(81x⁹y⁻¹⁰z⁶)]

= 9 * √(81/3 * x⁰x⁹ * y⁴y⁻¹⁰ * z⁶/z³)

= 9 * √(27 x⁹ y⁻⁶ z³)

= 9 * √(9x⁸y⁻⁶z² * 3xz)

= 9 * 3x⁴y⁻³z * √(3xz)

= 27x⁴z/y³ * √(3xz)

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Case 2: cube root

³√(x⁰y⁴/3z³) * ³√(81x⁹y⁻¹⁰z⁶)

= ³√[(x⁰y⁴/3z³)*(81x⁹y⁻¹⁰z⁶)]

= ³√(81/3 * x⁰x⁹ * y⁴y⁻¹⁰ * z⁶/z³)

= ³√(27 x⁹ y⁻⁶ z³)

= 3 x³ y⁻² z

= 3 x³ z / y²

http://mathway.com/answer.aspx?p=calg?p=3SMB20xSMB...