Question 132

If $$x-\frac{1}{x}=1$$, then the value of $$\frac{x^{4}-\frac{1}{x^{2}}}{3x^{2}+5x-3}$$ is

Solution

In the expression, $$\frac{x^{4}-\frac{1}{x^{2}}}{3x^{2}+5x-3}$$

Dividing numerator and denominator by $$x$$, we get

= $$\frac{x^{3}-\frac{1}{x^{3}}}{3x+5-\frac{3}{x}}$$ = $$\frac{x^{3}-\frac{1}{x^{3}}}{3(x-\frac{1}{x})+5}$$

= $$\frac{(x-\frac{1}{x})^{3} + 3(x-\frac{1}{x})}{3(x-\frac{1}{x})+5}$$

Now, putting $$x-\frac{1}{x}$$ = 1

we get, = $$\frac{1+3}{3+5} = \frac{4}{8} = \frac{1}{2}$$

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