If $$x + \frac{1}{x} = 3,$$ x ≠ 0 then the value of $${x^7} + \frac{1}{{{x^7}}}$$ is
A. 746
B. 843
C. 749
D. 849
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{x} = 3,\,x \ne 0 \cr & {x^2} + \frac{1}{{{x^2}}} = {3^2} - 2 = 7 \cr & {x^4} + \frac{1}{{{x^4}}} = {7^2} - 2 = 47 \cr & {x^3} + \frac{1}{{{x^3}}} = {3^3} - 3 \times 3 \cr & = 27 - 9 \cr & = 18 \cr & {x^7} + \frac{1}{{{x^7}}} = \left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) - \left( {x + \frac{1}{x}} \right) \cr & = 47 \times 18 - 3 \cr & = 846 - 3 \cr & = 843 \cr} $$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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