Question

x3+y3+z3-3xyz=1/2(x+y+z)[(x-ysquare)+(y-z)square+(z-xsquare)]

Answer 1

$<b><i>Hey mate!!$

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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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Thanks for the question!

Answer 2

here is your answer

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)]