If xy = -6 and x3 + y3 = 19 (x and y are integers), then what is the value of $$\frac{1}{{{x^{ - 1}}}} + \frac{1}{{{y^{ - 1}}}}?$$
A. 2
B. 1
C. -2
D. -1
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & xy = - 6,\,{x^3} + {y^3} = 19 \cr & {\text{putting }}x = 3,\,y = - 2 \cr & {\text{both condition are satisfied}} \cr & {\text{So, }}\frac{1}{{{x^{ - 1}}}} + \frac{1}{{{y^{ - 1}}}} \cr & = x + y \cr & = 3 - 2 \cr & = 1 \cr} $$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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