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