Ex 9.5

Chapter 9 Class 8 Algebraic Expressions and Identities
Serial order wise

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Ex 9.5, 5 Show that. (iv) (4ππ+3π)^2β(4ππβ3π)^2=48ππ^2 Solving (πππ+ππ)^π (π+π)^2=π^2+π^2+2ππ Putting π = 4ππ & π = 3π (π+π)^2=π^2+π^2+2ππ Putting π = 4ππ & π = 3π = (4ππ)^2+(3π)^2+2(4ππ)(3π) = 16π^2 π^2+9π^2+24ππ^2 Solving (πππβππ)^π (πβπ)^2=π^2+π^2β2ππ Putting π = 4ππ & π = 3π = (4ππ)^2+(3π)^2β2(4ππ)(3π) = 16π^2 π^2+9π^2β24ππ^2 Solving LHS (4ππ+3π)^2β(4ππβ3π)^2 = (16π^2 π^2+9π^2+24ππ^2 )β(16π^2 π^2+9π^2β24ππ^2 ) = 16π^2 π^2+9π^2+24ππ^2β16π^2 π^2β9π^2+24ππ^2 = (16π^2 π^2β16π^2 π^2 )+(9π^2β9π^2 )+(24ππ^2+24ππ^2 ) = 0+0+48ππ^2 = 48ππ^2 = R.H.S Since LHS = RHS Hence proved