Respuesta :

kudzordzifrancis

Answer:

[tex]32 {x}^{2} \sqrt[3]{ 2x } -8{x}^{3}[/tex]

Step-by-step explanation:

We want to

[tex]4x \sqrt[3]{4 {x}^{2} } (2 \sqrt[3]{32 {x}^{2} } - x \sqrt[3]{2x} )[/tex]

We expand to obtain:

[tex]4x \sqrt[3]{4 {x}^{2} } \times 2 \sqrt[3]{32 {x}^{2} } -4x \sqrt[3]{4 {x}^{2} } \times x \sqrt[3]{2x} )[/tex]

We now simplify

[tex]8x \sqrt[3]{4 {x}^{2} \times 32 {x}^{2} } -4 {x}^{2} \sqrt[3]{4 {x}^{2} \times 2x} [/tex]

We multiply the radicand

[tex]8x \sqrt[3]{64 \times {x}^{3} \times 2x } -4 {x}^{2} \sqrt[3]{8 {x}^{3}} [/tex]

Or

[tex]8x \sqrt[3]{ {(4x)}^{3} \times 2x } -4 {x}^{2} \sqrt[3]{{(2x)}^{3}} [/tex]

We take cube root to get:

[tex]8x \times 4x\sqrt[3]{ 2x } -4 {x}^{2} \times 2x[/tex]

We multiply out to get:

[tex]32 {x}^{2} \sqrt[3]{ 2x } -8{x}^{3}[/tex]

cerverusdante

Answer:

[tex]32x^2\sqrt[3]{2x} -8x^3[/tex]

Step-by-step explanation:

To multiply radicals of the same index, we multiply the coefficient of the radicals together and the radicands (the things inside the radical) together.

[tex]4x\sqrt[3]{4x^2} (2\sqrt[3]{32x^2} -x\sqrt[3]{2x} )[/tex]

[tex](4x)(2)\sqrt[3]{(4x^2)(32x^2)} -(4x)(x)\sqrt[3]{(4x^2)(2x)}[/tex]

[tex]8x\sqrt[3]{128x^4} -4x^2\sqrt[3]{8x^3}[/tex]

Remember that [tex]128=2^7=(2^6)(2)[/tex], [tex]8=2^3[/tex], and [tex]x^4=(x^3)(x)[/tex], so:

[tex]8x\sqrt[3]{(2^6)(2)(x^3)(x)} -4x^2\sqrt[3]{2^3x^3}[/tex]

Remember that radicands with the same index (or evenly divisible by the index) can be taken out the radical, so:

[tex](2^2)(x)(8x)\sqrt[3]2{x} -(4x^2)(2x)[/tex]

[tex]32x^2\sqrt[3]{2x} -8x^3[/tex]

We can conclude that the second choice is the correct answer.

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