If $$x = {3^{\frac{1}{3}}} - {3^{ - \frac{1}{3}}}$$ value of $$3{x^3} + 9x$$ is?
A. 8
B. 9
C. 27
D. 16
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & x = {3^{\frac{1}{3}}} - {3^{ - \frac{1}{3}}} \cr & \Rightarrow x + {3^{ - \frac{1}{3}}} = {3^{\frac{1}{3}}}\,\,\left( {{\text{cubing both sides}}} \right) \cr & \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}}\left( {x + {3^{ - \frac{1}{3}}}} \right) = 3 \cr & \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}} \times {3^{\frac{1}{3}}} = 3 \cr & \,\,\,\,\,\,\,\,\,\left( {{\text{multiply both sides by 3}}} \right) \cr & \Rightarrow 3{x^3} + 9x = 9 - 1 \cr & \Rightarrow 3{x^3} + 9x = 8 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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