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Question Number 162027 by mathlove last updated on 25/Dec/21

Answered by MJS_new last updated on 25/Dec/21

x∈R∧y∈R∧n∈Z: y=((−x))^(1/(2n+1)) =−(x)^(1/(2n+1)) ⇒ ((−9^(−2) ))^(1/3) =−(9^(−2) )^(1/3) =−((1/9^2 ))^(1/3) =−(1/( ((81))^(1/3) ))=−(1/(3(3)^(1/3) ))

x∈R∧y∈R∧n∈Z:y=−x2n+1=−x2n+1⇒−9−23=−9−23=−1923=−1813=−1333

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