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find square root


6 years ago


Answers : (1)


Hello Ajinkya,

lets say (x-y)4+2(x4+y4)-2(x2+y2)(x-y)2 as 'A'

A= (x-y)4+2(x4+y4)-2(x2+y2)(x-y)2

Taking the  (x-y)2 common from 1st term and the 3 rd term in 'A' , we have,

A = 2(x4+y4) + (x-y)2((x-y)2-2(x2+y2))

We know that 2(x2+y2) = (x-y)2 + (x+y)2, on substituting this in 'A', we have

A = 2(x4+y4) +(x-y)2((x-y)2-(x-y)2-(x+y)2) , cancelling out + and - (x-y)2 ,we get

A = 2(x4+y4) -(x-y)2(x+y)2

we know that (a-b)*(a+b) = a2-b2 ; so (x-y)2(x+y)2 = (x2-y2)2

therefore, we are left with A = 2(x4+y4) - (x2-y2)2

Again we know that (x2-y2)2 + (x2+y2)2= 2(x4+y4)

on replaching 2(x4+y4) in A with above terms we get ,

A=(x2-y2)2 + (x2+y2)2-(x2-y2)2

We get A = (x2+y2)2

We are asked to find the sqrt of A , which will be x2+y2.


Best of luck ajinkya.

You can ask us any questions as we are dedicated IITians group specially here to answer your queries.


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6 years ago

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