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bhavesh

asked 2016-05-26 08:19:09

if $x=a+\dfrac{1}{a}$ and $y=a-\dfrac{1}{a}$, find $x^4+y^4-2x^2y^2$

Javed Khan

answered 2016-05-29 13:15:02

$x^2-y^2\\

=\left(a+\dfrac{1}{a}\right)^2-\left(a-\dfrac{1}{a}\right)^2\\

=\left(a^2+2+\dfrac{1}{a^2}\right)-\left(a^2-2+\dfrac{1}{a^2}\right)\\

=4$

$\left(x^2-y^2\right)^2=4^2\\

x^4+y^4-2x^2y^2=16$

=\left(a+\dfrac{1}{a}\right)^2-\left(a-\dfrac{1}{a}\right)^2\\

=\left(a^2+2+\dfrac{1}{a^2}\right)-\left(a^2-2+\dfrac{1}{a^2}\right)\\

=4$

$\left(x^2-y^2\right)^2=4^2\\

x^4+y^4-2x^2y^2=16$

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