Question

Sea $u = \left( -\frac{1}{27} \right)$. Entonces: $u - \left( -\frac{1}{27} \right)^{\frac{2}{3}} = \left[ \left( -\frac{1}{27} \right)^{\frac{1}{3}} \right]^2$ $= \left[ \left( -\frac{1}{3^3} \right)^{\frac{1}{3}} \right]^2$ $= \left( -\frac{1}{3} \right)^2 = \frac{1}{9}$ $\frac{1}{3} - \frac{1}{3} - \frac{1}{3} - \frac{1}{243}$

          Sea $u = \left( -\frac{1}{27} \right)$. Entonces:
$u - \left( -\frac{1}{27} \right)^{\frac{2}{3}} = \left[ \left( -\frac{1}{27} \right)^{\frac{1}{3}} \right]^2$
$= \left[ \left( -\frac{1}{3^3} \right)^{\frac{1}{3}} \right]^2$
$= \left( -\frac{1}{3} \right)^2 = \frac{1}{9}$
$\frac{1}{3} - \frac{1}{3} - \frac{1}{3} - \frac{1}{243}$

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Sea u = ( -(1)/(27)). Entonces:
u - ( -(1)/(27))^(2)/(3) = [ ( -(1)/(27))^(1)/(3)]^2
= [ ( -(1)/(3^3))^(1)/(3)]^2
= ( -(1)/(3))^2 = (1)/(9)
(1)/(3) - (1)/(3) - (1)/(3) - (1)/(243)

Added by Kimberly A.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

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Solved by Expert Madhur L

07/12/2022

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First, let's break down the given expression: "Sea u = Enronces: ~OJ[) Jac 0 24}" Show more…

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