If t2 - 4t + 1 = 0, then the value of $${t^3} + \frac{1}{{{t^3}}}$$ is?
A. 44
B. 48
C. 52
D. 64
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {t^2} - 4t + 1 = 0 \cr & \Rightarrow {t^2} + 1 = 4t \cr & \Rightarrow \frac{{{t^2} + 1}}{t} = \frac{{4t}}{t} \cr & \Rightarrow t + \frac{1}{t} = 4 \cr & \left[ {{\text{Take cube both sides}}} \right] \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3.t.\frac{1}{t}\left( {t + \frac{1}{t}} \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3\left( 4 \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 64 - 12 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 52 \cr} $$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. $$\frac{8}{6}$$
Join The Discussion