Question

If x-1/x = 2, then x⁴-1/x⁴ =?

Answer 1

$\textbf{Given:}$

$x-\dfrac{1}{x}=2$

$\textbf{To find:}$

$x^4-\dfrac{1}{x^4}$

$\textbf{Solution:}$

$\text{Consider,}$

$x-\dfrac{1}{x}=2$

$\text{Squaring on bothsides, we get}$

$(x-\dfrac{1}{x})^2=4$

$x^2+\dfrac{1}{x^2}-2=4$

$x^2+\dfrac{1}{x^2}=4+2$

$\implies\bf\,x^2+\dfrac{1}{x^2}=6$

$(x+\dfrac{1}{x})^2=x^2+\dfrac{1}{x^2}+2(x)(\dfrac{1}{x})$

$(x+\dfrac{1}{x})^2=6+2$

$(x+\dfrac{1}{x})^2=8$

$\implies\bf\,x+\dfrac{1}{x}=2\sqrt{2}$

$\text{Now,}$

$x^4-\dfrac{1}{x^4}$

$=(x^2)^2-(\dfrac{1}{x^2})^2$

$\text{Using,}\;\boxed{\bf\,a^2-b^2=(a-b)(a+b)}$

$=(x^2-\dfrac{1}{x^2})(x^2+\dfrac{1}{x^2})$

$=(x-\dfrac{1}{x})(x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2})$

$=(2)(2\sqrt{2})(6)$

$=24\,\sqrt{2}$

$\therefore\bf\,x^4-\dfrac{1}{x^4}=24\,\sqrt{2}$

Find more:

Ax+by=1 bx+ay=2ab/a2+b2 then (x*2+y*2) (a*2+b*2)=?

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