If x = a^{1/2} + a^{-1/2} and y = a^{1/2} - a | vedclass

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(C) Given: $x = a^{1/2} + a^{-1/2}$ and $y = a^{1/2} - a^{-1/2}$.First,calculate $x^2$ and $y^2$:$x^2 = (a^{1/2} + a^{-1/2})^2 = a + a^{-1} + 2(a^{1/2

Vedclass · Accepted Answer

$16$

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(C) Given: $x = a^{1/2} + a^{-1/2}$ and $y = a^{1/2} - a^{-1/2}$.First,calculate $x^2$ and $y^2$:$x^2 = (a^{1/2} + a^{-1/2})^2 = a + a^{-1} + 2(a^{1/2})(a^{-1/2}) = a + a^{-1} + 2$$y^2 = (a^{1/2} - a^{-1/2})^2 = a + a^{-1} - 2(a^{1/2})(a^{-1/2}) = a + a^{-1} - 2$Now,calculate $x^2 - y^2$:$x^2 - y^2 = (a + a^{-1} + 2) - (a + a^{-1} - 2) = 4$Also,calculate $x^2 y^2$:$x^2 y^2 = (a + a^{-1} + 2)(a + a^{-1} - 2) = (a + a^{-1})^2 - 2^2 = (a + a^{-1})^2 - 4$The expression is $(x^4 - x^2 y^2 - 1) + (y^4 - x^2 y^2 + 1) = x^4 + y^4 - 2x^2 y^2$.This simplifies to $(x^2 - y^2)^2$.Substituting the value $x^2 - y^2 = 4$:$(x^2 - y^2)^2 = 4^2 = 16$.

If $x = a^{1/2} + a^{-1/2}$ and $y = a^{1/2} - a^{-1/2}$,then find the value of $(x^4 - x^2 y^2 - 1) + (y^4 - x^2 y^2 + 1)$.

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