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If x2 - 5x - 1 = 0, what is the value of $\frac{x^6-x^4+x^2-1}{x^3}$ ? |
145 140 130 135 |
135 |
If x2 - 5x - 1 = 0 x - \(\frac{1}{x}\) = 5 x3 - \(\frac{1}{x^3}\) = 53 + 3 × 5 = 140 $\frac{x^6-x^4+x^2-1}{x^3}$ = x3 - \(\frac{1}{x^3}\) - (x - \(\frac{1}{x}\)) = 140 - 5 = 135 |
