Question 116

If $$ x^{2}-3x+1=0$$ then the value of $$\frac{x^{6}+x^{4}+x^{2}+1}{x^3}$$ will be

Solution

Expression : $$ x^{2}-3x+1=0$$

=> $$x^2 + 1 = 3x$$

=> $$x + \frac{1}{x} = 3$$ ------------------Eqn(1)

To find : $$\frac{x^{6}+x^{4}+x^{2}+1}{x^3}$$

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

= $$[(x + \frac{1}{x})^3 - 3.x.\frac{1}{x}(x + \frac{1}{x})] + (x + \frac{1}{x})$$

Using eqn (1)

= $$3^3 -3*3 + 3$$ = 27-9+3

= 21

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