If $${x^3} + \frac{3}{x}$$ = $$4\left( {{a^3} + {b^3}} \right)$$ and $$3x + \frac{1}{{{x^3}}}$$ = $$4\left( {{a^3} - {b^3}} \right){\text{,}}$$ then a2 - b2 is equal to?
A. 4
B. 0
C. 1
D. 2
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {x^3} + \frac{3}{x} = 4\left( {{a^3} + {b^3}} \right)\, . . . . . {\text{(i)}} \cr & 3x + \frac{1}{{{x^3}}} = 4\left( {{a^3} - {b^3}} \right)\, . . . . . (ii) \cr & {\text{Equation (i)}} + {\text{(ii)}} \cr & {\left( {x + \frac{1}{x}} \right)^3} = 8{a^3} \cr & \Leftrightarrow x + \frac{1}{x} = 2a\, . . . . . (iii) \cr & Equation{\text{ }}(i) - (ii) \cr & x - \frac{1}{x} = 2b\, . . . . . (iv) \cr & Equation{\text{ }}(iii) - (iv) \cr & 2\left( {a - b} \right) = \frac{2}{x} \cr & \Rightarrow a - b = \frac{1}{x} \cr & \Rightarrow a + b = x \cr & \because {a^2} - {b^2} = \left( {a + b} \right) \times \left( {a - b} \right) \cr & \Rightarrow {a^2} - {b^2} = x \times \frac{1}{x} \cr & \Rightarrow {a^2} - {b^2} = 1 \cr} $$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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