If x+frac{1}{x}=5, then x^{6}+frac{1}{x^{6}} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x+\frac{1}{x}=5$.First,find $x^3+\frac{1}{x^3}$ by cubing both sides:$(x+\frac{1}{x})^3 = 5^3$$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{

Vedclass · Accepted Answer

$12098$

Vedclass · Answer

(A) Given $x+\frac{1}{x}=5$.First,find $x^3+\frac{1}{x^3}$ by cubing both sides:$(x+\frac{1}{x})^3 = 5^3$$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x}) = 125$$x^3+\frac{1}{x^3}+3(5) = 125$$x^3+\frac{1}{x^3} = 125-15 = 110$.Now,square both sides to find $x^6+\frac{1}{x^6}$:$(x^3+\frac{1}{x^3})^2 = 110^2$$x^6+\frac{1}{x^6}+2(x^3)(\frac{1}{x^3}) = 12100$$x^6+\frac{1}{x^6}+2 = 12100$$x^6+\frac{1}{x^6} = 12100-2 = 12098$.

If $x+\frac{1}{x}=5,$ then $x^{6}+\frac{1}{x^{6}}$ is

Similar Questions

If $a + b + c = 0$,then the roots of the equation $4ax^2 + 3bx + 2c = 0$ are

If $(1 - p)$ is a root of the quadratic equation $x^2 + px + (1 - p) = 0$,then its roots are:

$I. x^{2}-20 x+91=0$
$II. y^{2}-32 y+247=0$

The equation $x^{(3/4)(\log_2 x)^2 + (\log_2 x) - 5/4} = \sqrt{2}$ has

Solve the given equations and select the correct option.
$I.$ $7x + 6y = 110$
$II.$ $4x + 3y = 59$
$III.$ $x + z = 15$

Vedclass Test Series

Exam Paper Generator

Online Exam Module