If x=\sqrta+1/\sqrta,y=\sqrta-1/\sqrta,(a>0), then the value of x^4+y^4-2x^2y^2 is

RRB Junior Engineer Stage 1 Model Papers 3

Question : 10 of 100

Marks: +1, -0

x=√a+1/√a,y=√a-1/√a,(a>0),

then the value of

x^4+y^4-2x^2y^2

Solution:

x=√a+1/√a \;y=√a-1/√a

x^4+y^4-2x^2y^2

=(x^2-y^2)^2

x^2=a+1/a+2

y^2=a+1/a-2

(x^2-y^2)^2=[a+1/a+2-a-1/a+2]^2=(4)^2=16

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