Math, asked by Ritul33, 1 year ago

Verify that : x³ + y³ + z³− 3xyz =

1/2 (x + y + z)[(x − y)² + (y − z)² + (z − x)2²]

Answers

Answered by srikanthn711

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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)]

$$

Hope this will help you..

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Answered by Anonymous

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)]

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