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Question Number 18667 by chernoaguero@gmail.com last updated on 27/Jul/17

$$\mathrm{If}x=2^{1/3}-2^{-1/3},\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$$2x^3+6x.$$

Answered by ajfour last updated on 27/Jul/17

$${2\mathrm{x}}^3=2[2-3(2^{1/3}-2^{-1/3})-\frac{1}{2}]$$$$=3-6\mathrm{x}$$$$\Rightarrow {2\mathrm{x}}^3+6\mathrm{x}=3.$$

Answered by Arnab Maiti last updated on 27/Jul/17

$$2x^3+6x=2{(2^{\frac{1}{3}}-2^{-\frac{1}{3}})}^3+6(2^{\frac{1}{3}}-2^{\frac{-1}{3}})$$$$=2\times [2-3\times 2^{\frac{2}{3}}\times 2^{\frac{-1}{3}}+3\times 2^{\frac{1}{3}}\times 2^{\frac{-2}{3}}-2^{-1}]+6(2^{\frac{1}{3}}-2^{\frac{-1}{3}})$$$$=2\times [\{2-\frac{1}{2}\}-3\{2^{\frac{1}{3}}-2^{\frac{-1}{3}}\}]+6(2^{\frac{1}{3}}-2^{\frac{-1}{3}})$$$$=2\times \left\{\frac{4-1}{2}\right\}-6\times \{2^{\frac{1}{3}}-2^{\frac{-1}{3}}\}+6\times (2^{\frac{1}{3}}-2^{\frac{-1}{3}})$$$$=3$$

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