If x^{4}+\frac{1}{x^4}=119 and x<1 the find the positive value of x^{3}-\frac{1}{x^3}

Question 125

If $$x^{4}+\frac{1}{x^4}=119$$ and $$x<1$$ the find the positive value of $$x^{3}-\frac{1}{x^3}$$

Solution

Expression : $$x^{4}+\frac{1}{x^4}=119$$

=> $$x^4 + \frac{1}{x^4} + 2 = 121$$

=> $$(x^2 + \frac{1}{x^2})^2 = 11^2$$

=> $$x^2 + \frac{1}{x^2} = 11$$

=> $$x^2 + \frac{1}{x^2} - 2 = 9$$

=> $$(x - \frac{1}{x})^2 = 3^2$$

=> $$x - \frac{1}{x} = 3$$

Now, cubing both sides, we get :

=> $$x^3 - \frac{1}{x^3} - 3.x.\frac{1}{x}.(x-\frac{1}{x}) = 27$$

=> $$x^3 - \frac{1}{x^3} = 27+3*3$$ = 36

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