find square root(x-y)4+2(x4+y4)-2(x2+y2)(x-y)2

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) = a^₂-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.

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