If 4x2 - 6x + 1 = 0, then the value of $$8{x^3} + \frac{1}{{8{x^3}}}$$ is:
A. 36
B. 13
C. 18
D. 11
Answer: Option C
Solution (By Examveda Team)
$$\eqalign{ & 4{x^2} - 6x + 1 = 0 \cr & 2x\left( {2x - 3 + \frac{1}{{2x}}} \right) = 0 \cr & 2x + \frac{1}{{2x}} = 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 27 - 3 \times 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 18 \cr} $$Related Questions on Algebra
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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