If x = sqrt[3]{{sqrt 2 + 1}} - sqrt[3]{{sqrt | vedclass

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(A) Given $x = {(\sqrt 2 + 1)^{1/3}} - {(\sqrt 2 - 1)^{1/3}}$.Using the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$,we have:${x^3} = (\sqrt 2 + 1) -

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$2$

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(A) Given $x = {(\sqrt 2 + 1)^{1/3}} - {(\sqrt 2 - 1)^{1/3}}$.Using the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$,we have:${x^3} = (\sqrt 2 + 1) - (\sqrt 2 - 1) - 3{(\sqrt 2 + 1)^{1/3}}{(\sqrt 2 - 1)^{1/3}} \left[ {(\sqrt 2 + 1)^{1/3}} - {(\sqrt 2 - 1)^{1/3}} \right]$.Since ${(\sqrt 2 + 1)^{1/3}}{(\sqrt 2 - 1)^{1/3}} = {((\sqrt 2 + 1)(\sqrt 2 - 1))^{1/3}} = {(2 - 1)^{1/3}} = 1$,the equation becomes:${x^3} = 2 - 3(1)x$.Therefore,${x^3} + 3x = 2$.

If $x = \sqrt[3]{{\sqrt 2 + 1}} - \sqrt[3]{{\sqrt 2 - 1}}$,then find the value of ${x^3} + 3x$.

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