If $$x - \frac{1}{x} = 1{\text{,}}$$ then the value of $$\frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}}$$ = ?
A. $$\frac{1}{4}$$
B. $$\frac{1}{2}$$
C. $$\frac{3}{4}$$
D. 0
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & x - \frac{1}{x} = 1{\text{ }} \cr & \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr & {\text{Divide and multiply by }}x \cr & \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr & \Rightarrow \frac{4}{8} \cr & \Rightarrow \frac{1}{2} \cr} $$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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