If x = {2^{1/3}} - {2^{ - 1/3}}, then 2{x^3} | vedclass

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Question

(c) $x = {2^{1/3}} - {2^{ - 1/3}}$

$ \Rightarrow $${x^3} = 2 - {2^{ - 1}} - {3.2^{1/3}}{.2^{ - 1/3}}({2^{1/3}} - {2^{ - 1/3}})$

$ \Rightarrow $$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(c) $x = {2^{1/3}} - {2^{ - 1/3}}$

$ \Rightarrow $${x^3} = 2 - {2^{ - 1}} - {3.2^{1/3}}{.2^{ - 1/3}}({2^{1/3}} - {2^{ - 1/3}})$

$ \Rightarrow $${x^3} = 2 - {1 \over 2} - 3x$ ==> ${x^3} + 3x = {3 \over 2}$

$	herefore 2{x^3} + 6x = 3$.

If $x = {2^{1/3}} - {2^{ - 1/3}},$ then $2{x^3} + 6x = $

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