If $${x^2} - 3x + 1 = 0,$$ then the value of $${x^3}{\text{ + }}\frac{1}{{{x^3}}}\,{\text{is?}}$$
A. 9
B. 18
C. 27
D. 1
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & {\text{Divide by }}x \cr & \frac{{{x^2}}}{x} + \frac{1}{x} = \frac{{3x}}{x} \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 18 \cr} $$Join The Discussion
Comments ( 1 )
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. $$\frac{8}{6}$$
solution is not matching with the question