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VladimirAG [237]
3 years ago
6

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

Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
6 0

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}

Molodets [167]3 years ago
3 0

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.

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