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\[\sqrt[3]{x ^{2}}\times \sqrt[3]{x ^{2}}\]
\[\Large \sqrt[n]{x^m} = x^\frac{ m }{ n} \quad \]
\[\Large x^m * x^n = x^{(n+m)}\]
\[x ^{\frac{2}{3}}\times x ^{\frac{2}{3}}\]
So, would the answer be \[\sqrt[3]{x}\]
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You need to add the exponents: x^(2/3) * x^(2/3) = x^(2/3 + 2/3) = ?
So, it'd be x^4/6, but you have to simplify, so x^2/3
2/3 + 2/3 is not 4/6
Oh woops! 4/3
Yes. x^(4/3) which can be put back in the radical form as:\[\Large \sqrt[3]{x^4}\]
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a.\[x \sqrt[3]{x}\] b. \[\sqrt[3]{x}\] c. x d. \[x ^{\frac{ 4 }{ 9 }}\]
I'm confused, because those are my options.
\[\Large \sqrt[3]{x^4} = \sqrt[3]{x^3 * x} = \sqrt[3]{x^3} * \sqrt[3]{x} = x * \sqrt[3]{x}\]
Oh! I see! Thank you!
you are welcome.
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