If $$x + \frac{1}{{16x}} = 3,$$ then the value of $$16{x^3} + \frac{1}{{256{x^3}}}$$ is:
A. 423
B. 441
C. 432
D. 414
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{{16x}} = 3 \cr & 2x + \frac{1}{{8x}} = 6 \cr & {\text{Cube both side}} \cr & 8{x^3} + \frac{1}{{512{x^3}}} + 3 \times 2 \times \frac{1}{8} \times 6 = 216 \cr & 8{x^3} + \frac{1}{{512{x^3}}} = 216 - \frac{9}{2} \cr & {\text{Multiply by '2' both side}} \cr & 16{x^3} + \frac{1}{{256{x^3}}} = 432 - 9 = 423 \cr} $$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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