# Show that(i) is purely real,(ii) is purely real.

Given:

Taking the L.C.M, we get

[(a + b)(a – b) = (a2 – b2)]

Putting i2 = -1

= 0 + 0i

Hence, the given equation is purely real as there is no imaginary part.

(ii) Given:

Taking the L.C.M, we get

…(i)

[(a + b)(a – b) = (a2 – b2)]

Now, we know that,

(a + b)2 + (a – b)2 = 2(a2 + b2)

So, by applying the formula in eq. (i), we get

Putting i2 = -1

Hence, the given equation is purely real as there is no imaginary part.

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