Question
If $$x = {3^{\frac{1}{3}}} - {3^{ - \frac{1}{3}}}$$ value of $$3{x^3} + 9x$$ is?
Answer: Option A
$$\eqalign{
& x = {3^{\frac{1}{3}}} - {3^{ - \frac{1}{3}}} \cr
& \Rightarrow x + {3^{ - \frac{1}{3}}} = {3^{\frac{1}{3}}}\,\,\left( {{\text{cubing both sides}}} \right) \cr
& \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}}\left( {x + {3^{ - \frac{1}{3}}}} \right) = 3 \cr
& \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}} \times {3^{\frac{1}{3}}} = 3 \cr
& \,\,\,\,\,\,\,\,\,\left( {{\text{multiply both sides by 3}}} \right) \cr
& \Rightarrow 3{x^3} + 9x = 9 - 1 \cr
& \Rightarrow 3{x^3} + 9x = 8 \cr} $$
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$$\eqalign{
& x = {3^{\frac{1}{3}}} - {3^{ - \frac{1}{3}}} \cr
& \Rightarrow x + {3^{ - \frac{1}{3}}} = {3^{\frac{1}{3}}}\,\,\left( {{\text{cubing both sides}}} \right) \cr
& \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}}\left( {x + {3^{ - \frac{1}{3}}}} \right) = 3 \cr
& \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}} \times {3^{\frac{1}{3}}} = 3 \cr
& \,\,\,\,\,\,\,\,\,\left( {{\text{multiply both sides by 3}}} \right) \cr
& \Rightarrow 3{x^3} + 9x = 9 - 1 \cr
& \Rightarrow 3{x^3} + 9x = 8 \cr} $$
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