which expression is equivalent to $sqrt3{256x^{10}y^{7}}$?\n$4x^{2}y(sqrt3{x^{2}y^{3}})$\n$4x^{3}y^{2}(sqrt3{…

Explanation:

Step1: Prime - factorize 256

$256 = 2^8$. So, $\sqrt[3]{256x^{10}y^{7}}=\sqrt[3]{2^8x^{10}y^{7}}$.

Step2: Use the property $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$

$\sqrt[3]{2^8x^{10}y^{7}}=\sqrt[3]{2^8}\cdot\sqrt[3]{x^{10}}\cdot\sqrt[3]{y^{7}}$.

Step3: Rewrite exponents in terms of multiples of 3

$2^8 = 2^{6 + 2}$, $x^{10}=x^{9+1}$, $y^{7}=y^{6 + 1}$. Then $\sqrt[3]{2^8}\cdot\sqrt[3]{x^{10}}\cdot\sqrt[3]{y^{7}}=\sqrt[3]{2^{6}\cdot2^{2}}\cdot\sqrt[3]{x^{9}\cdot x^{1}}\cdot\sqrt[3]{y^{6}\cdot y^{1}}$.

Step4: Use the property $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$ again

$\sqrt[3]{2^{6}\cdot2^{2}}\cdot\sqrt[3]{x^{9}\cdot x^{1}}\cdot\sqrt[3]{y^{6}\cdot y^{1}}=\sqrt[3]{2^{6}}\cdot\sqrt[3]{2^{2}}\cdot\sqrt[3]{x^{9}}\cdot\sqrt[3]{x^{1}}\cdot\sqrt[3]{y^{6}}\cdot\sqrt[3]{y^{1}}$.

Step5: Simplify cube - roots

$\sqrt[3]{2^{6}} = 2^{2}=4$, $\sqrt[3]{x^{9}}=x^{3}$, $\sqrt[3]{y^{6}}=y^{2}$. So, $\sqrt[3]{2^{6}}\cdot\sqrt[3]{2^{2}}\cdot\sqrt[3]{x^{9}}\cdot\sqrt[3]{x^{1}}\cdot\sqrt[3]{y^{6}}\cdot\sqrt[3]{y^{1}}=4\sqrt[3]{4}x^{3}y^{2}\sqrt[3]{xy}=4x^{3}y^{2}(\sqrt[3]{4xy})$.

Answer:

$4x^{3}y^{2}(\sqrt[3]{4xy})$