If x2 - 3x - 1 = 0, then the value of (x2 + 8x - 1)(x3 + x-1)-1 is:
A. $$\frac{3}{8}$$
B. 8
C. 1
D. 3
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {x^2} - 3x - 1 = 0 \cr & x\left( {x - 3 - \frac{1}{x}} \right) = 0 \cr & x - \frac{1}{x} = 3 \cr & {x^2} + \frac{1}{{{x^2}}} = 11 \cr & \frac{{\left( {{x^2} + 8x - 1} \right)}}{{{x^3} + \frac{1}{x}}} \cr & = \frac{{x\left( {\frac{{x - 1}}{{x + 8}}} \right)}}{{x\left( {\frac{{{x^2} + 1}}{{{x^2}}}} \right)}} \cr & = \frac{{3 + 8}}{{11}} \cr & = 1 \cr} $$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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