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How do I do this? 2y^1/3∙ 6y^1/3
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add the exponents
\[\large 2y^{1/3} \times 6y^{1/3}\]
also \(2\times 6=12\) so \[12y^{\frac{1}{3}+\frac{1}{3}}\]
So I would have \[12y ^{2/3}\]
yes
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