Question

Simplify the expression.

$\sqrt[3]{24x^{6}y^{2}}-2x^{2} \sqrt[3]{375y^{2}}-3x\sqrt[3]{16x^{3}y^{2}}$

Please help!

Answer 1

Step-by-step explanation:

several things to remember :

an nth root is the same as 1/n as exponent.

exponent of exponent means multiplying the exponents. like

(x⁴)³ = x¹²

when applying an exponent to a product of factors, then it has to be applied to each factor.

so the expression is first of all the same as

(24x⁶y²)^(1/3) - 2x²(375y²)^(1/3) - 3x(16x³y²)^(1/3)

and that is then

(3×8x⁶y²)^(1/3) - 2x²(3×125y²)^(1/3) - (2×8x³y²)^(1/3) =

= 2x²(3y²)^(1/3) - 10x²(3y²)^(1/3) - 2x(2y²)^(1/3) =

= -8x²(3y²)^(1/3) -2x(2y²)^(1/3) =

= -2x((4x(3y²)^(1/3) + (2y²)^(1/3)) =

= -2x((4x(3)^(1/3) + 2^(1/3))y^(2/3)

Answer 2

Given

• $\sqrt[3]{24x^6y^2}-2x^2\sqrt[3]{375y^2} -3x\sqrt[3]{16x^3y^2}$

Simplify

• $\sqrt[3]{24x^6y^2}-2x^2\sqrt[3]{375y^2} -3x\sqrt[3]{16x^3y^2} =$

• $\sqrt[3]{3*2^3(x^2)^3y^2}-2x^2\sqrt[3]{3*5^3*y^2} -3x\sqrt[3]{2*2^3*x^3y^2} =$

• $2x^2\sqrt[3]{3y^2} -10x^2\sqrt[3]{3y^2} -6x^2\sqrt[3]{2y^2} =$

• $-8x^2\sqrt[3]{3y^2} -6x^2\sqrt[3]{2y^2}$