Question

Multiply: 4x * root(3, 4x ^ 2) * (2 * root(3, 32x ^ 2) - x * root(3, 2x))

Answer 1

Answer:

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

Step-by-step explanation:

We want to

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

We expand to obtain:

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

We now simplify

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

We multiply the radicand

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

Or

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

We take cube root to get:

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

We multiply out to get:

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

Answer 2

Answer:

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

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.

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

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

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

Remember that $128=2^7=(2^6)(2)$, $8=2^3$, and $x^4=(x^3)(x)$, so:

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

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

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

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

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