Question

if a+b+c=0 then prove 1/x^b+x^-c+1+1/x^c+x^-a+1+1/x^a+x^-b+1=1if a+b+c=0 then prove 1/x^b+x^-c+1+1/x^c+x^-a+1+1/x^a+x^-b+1=1

Answer 1

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Answer :

The roots of the quadratic equation ,

$\frac{1}{x} - \frac{1}{x - 2} = 3 \: are \: x = \frac{3 + \sqrt[1]{3} }{3} \: and \: x = \frac{3 - \sqrt[1]{3} }{3}$ :

As given the equation in the form ,

$\frac{1}{x}-\frac{1}{x-2} = 3$

Simplify the above equation

=> (x-2)-x = 3x × (x-2)

=> x-2 - x = 3x² - 6x

=> 3x² - 6x + 2 = 0

As the equation is written in the form ax² + bx + c = 0

$x =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 3 , b = -6 , c = 2

Put all the values in the above equation

$x =\frac{-(-6)\pm\sqrt{(-6)^{2}-4\times 3\times 2}}{2\times 3}$

$x =\frac{6\pm\sqrt{36-24}}{6}$

$x =\frac{6\pm\sqrt{12}}{6}$

$x =\frac{6\pm2\sqrt{3}}{6}$

$x =\frac{3\pm1\sqrt{3}}{3}$

Thus,

$x =\frac{3+1\sqrt{3}}{3}$

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

Therefore the roots of the quadratic equation $\frac{1}{x}-\frac{1}{x-2} = 3$ are $x =\frac{3+1\sqrt{3}}{3}$,$x =\frac{3-1\sqrt{3}}{3}$

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