Question

good morning everyone
.12. Verify that x + y2 + z2 – 3xyz = + (x + y + z)[(x - y)² + (y – z)? + (z - x)2
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TC​​

Answer 1

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Answer 2

Solving R.H.S

1/2 (x+y+z)[(x - y)² + (y-z)² + (z − x)²]

Using (a - b)² = a² + b² - 2ab

1/2 (x+y+z) [ (x² + y² - 2xy) + (y² +z²-2yz) + (z² + x²-2zx)]

= 1/2 (x+y+z) [2x² + 2y² + 2z² - 2xy-2yz - 2zx]

=1/2 (x+y+z) 2 [x² + y² + z² − xy − yz - zx]

= (x+y+z) [x² + y² + z² - xy-yz-zx] We know x³ + y² + 2³-3xyz = (x+y+z) (x2 + y2 + 22-xy-yz - zx)

= x³ + y² + z³ - 3xyz

= L.H.S

L.H.S= R.H.S

HENCE PROVED