Will you square? Part 5

Note first that $\large 4 = 3 + 1 = (3^{\frac{1}{3}})^{3} + 1 = (3^{\frac{1}{3}} + 1)(3^{\frac{2}{3}} - 3^{\frac{1}{3}} + 1),$

where the identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$ has been used. Then

$\large \dfrac{4}{\sqrt[3]{9} - \sqrt[3]{3} + 1} = \dfrac{4}{3^{\frac{2}{3}} - 3^{\frac{1}{3}} + 1} = 3^{\frac{1}{3}} + 1 = (27^{\frac{1}{3}})^{\frac{1}{3}} + 1 = 27^{\frac{1}{9}} + 1.$

Thus $\large x = \dfrac{1}{9},$ and so $\large 81x = \boxed{9}.$

Substract 1 from both sides, and get: ${ 3 }^{ 3x }=\frac { 4-\sqrt [ 3 ]{ { 3 }^{ 2 } } +\sqrt [ 3 ]{ 3 } -1 }{ \sqrt [ 3 ]{ { 3 }^{ 2 } } -\sqrt [ 3 ]{ 3 } +1 } =\frac { \sqrt [ 3 ]{ { 3 }^{ 3 } } -\sqrt [ 3 ]{ { 3 }^{ 2 } } +\sqrt [ 3 ]{ 3 } }{ \sqrt [ 3 ]{ { 3 }^{ 2 } } -\sqrt [ 3 ]{ 3 } +1 } =\frac { \sqrt [ 3 ]{ 3 } \left( \sqrt [ 3 ]{ { 3 }^{ 2 } } -\sqrt [ 3 ]{ 3 } +1 \right) }{ \sqrt [ 3 ]{ { 3 }^{ 2 } } -\sqrt [ 3 ]{ 3 } +1 } =\sqrt [ 3 ]{ 3 } = { 3 }^{ \frac { 1 }{ 3 } }$

So 3x = $\frac{1}{3}$ , and if we multiply by 27, we get 81x = 9