Question

Verify that: x3+y3+z3–3xyz=(x+y+z)[(x–y)2+(y–z)2+(z–x)2]

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

First We take R.H.S & use the Formula [( a-b)²= a²+b²-2ab] & simplify it then R.H.S becomes equal to L.H.S

R.H.S

⇒ 1/2×(x + y + z) (x²+ y²-2xy +y²+ z²-2yz+x²+z²-2xz)

[( a-b)²= a²+b²-2ab]

⇒ 1/2×(x + y + z) (2x²+ 2y²+2z²-2xy -2yz-2xz)

⇒ 1/2×(x + y + z) 2(x² + y²+ z² – xy – yz – xz)

=(x + y + z) (x² + y²+ z² – xy – yz – xz)

= x³+y³+z³-3xyz

= L.H.S

We know that,

[x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)]

L.H.S = R.H.S

[x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)]

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