Question

Verify that

x^3 + y^3 + z^3 - 3xyz= 1/2 (x+y+z)( (x-y) ^2 + ( y-z)^2 + (z-x)^2 )

Answer 1

$\mathbb{ SOLUTION }$ :

$\bf {RHS}$
=> $\frac{1 }{2 }$ (x+y+z) { (x-y)² + (y-z)² + (z-x)² }

=> $\frac{x }{ 2}$ + $\frac{y }{2 }$ + $\frac{ z}{2 }$ [ x² - 2xy + y² + y² - 2yz + z² + z² - 2xz + x² ]

=> $\frac{x }{ 2}$ + $\frac{y }{2 }$ + $\frac{ z}{2 }$ [ 2x² + 2y² + 2z² - 2xy - 2yz - 2xz ]

=> $\frac{x }{ 2}$ [ 2x² + 2y² + 2z² - 2xy - 2yz - 2xz ] + $\frac{y }{2 }$ [ 2x² + 2y² + 2z² - 2xy - 2yz - 2xz ] + $\frac{ z}{2 }$ [ 2x² + 2y² + 2z² - 2xy - 2yz - 2xz ]

=> $\frac{x }{2}$ × 2x² + $\frac{x }{ 2}$ × 2y² + $\frac{x }{ 2}$ × 2z² - $\frac{x }{ 2}$ × 2xy - $\frac{x }{ 2}$ × 2yz - $\frac{x }{ 2}$ × 2xz + $\frac{y }{2 }$ × 2x² $\frac{y }{ 2}$ × 2y² + $\frac{y }{ 2}$ × 2z² - $\frac{y}{ 2}$ × 2xy - $\frac{y }{ 2}$ × 2yz - $\frac{y}{ 2}$ × 2xz + $\frac{z }{2}$ × 2x² + $\frac{z }{ 2}$ × 2y² + $\frac{z }{ 2}$ × 2z² - $\frac{z }{ 2}$ × 2xy - $\frac{z }{ 2}$ × 2yz - $\frac{z }{ 2}$ × 2xz

=> x³ + xy² + xz² - x²y - xyz - x²z + x²y + y³ + yz² - xy² - y²z - xyz + x²z - y²z + z³ - xyz - yz² - xz²

=> x³ + y³ + z³ - 3xyz = $\bf{ LHS}$

$\therefore$ $\bf{ RHS}$ = $\bf{ LHS}$

Verified.

Answer 2

  
Hey there !!


→ Prove that :-)

$\bf {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = \frac{1}{2} (x + y + z)( {(x - y) }^{2} + {(y - z)}^{2} + {(z - x)}^{2} ).$

→ solution :-)

Using Identity :-

=> x³ + y³ + z³ - 3xyz = ( x + y + z )( x² + y² + z² - xy - yz - zx ).

▶ We can write RHS in this form :-

$\bf = \frac{1}{2} (x + y + z)(2 {x}^{2} + 2 {y}^{2} + 2 {z}^{2} - 2xy - 2yz - 2zx).$

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➡ Because , $\bf = \frac{1}{\cancel{2}} (x + y + z) \times \cancel{2} ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx).$

= ( x + y + z )( x² + y² + z² -xy - yz - zx ).

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$\bf = \frac{1}{2} (x + y + z)( {x}^{2} + {x}^{2} + {y}^{2} + {y}^{2} + {z}^{2} + {z}^{2} - 2xy - 2yx - 2zx).$

= $\bf{ \frac{1}{2} }$ ( x + y + z ) [ ( x² - 2xy + y² ) + ( y² - 2yz + z² ) + ( z² - 2zx + x² ) ].

$\bf { {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = \frac{1}{2} (x + y + z)( {(x - y) }^{2} + {(y - z)}^{2} + {(z - x)}^{2} ). }$

✔✔ Hence, it is proved ✅✅.

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$\huge \boxed{ \mathbb{THANKS}}$

$\huge \bf{ \#BeBrainly.}$