1.

If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b c prove that a2 + b2 + c2 + 2abc = 1

Answer»

If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b c

x2 (y + z) y2 (z + x) z2 (x + y) = a2b2c2

x2y2z2 (x + y) (y + z) (z + x) = a2b2c2

xyz = abc

(x + y) (y + z) (z + x) = 1

a2 + b2 + c2 + 2abc 

=> x2(y + z) + y2 (z + x) + z2(x + y) + 2xyz

Put y = - z

0 + z2(z + x) + z2(x - z) + 2x (-z)z

z3 + z2x + z2x - z3 - 2z2x = 0

∴ (y + z) is factors

Similarly (x + y) & (x + z) are factors

x2(y + z) + y2(z + x) + z2(x + y) + 2xyz

= k (x + y) (y + z) (x + z)

For k put x = 0    y = 1     z = 1

0 + 1 + 1 + 0 = k (1) (2) (1)

2 = 2 k

k = 1

x2(y + z) + y2(z + x) + z2(x + y) + 2xyz

= (x + y) (y + z) (z + x)

and (x + y) (y + z) (z + x) = 1

∴ x2(y + z) + y2(z + x) + z2 (x + y) + 2xyz = 1


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