Question

prove that x^3+y^3+z^3-3xyz={(x+,y+z)[(x-y)^2+(y-z)^2+(z-x)^2]}/2..

plz answer fast..it's urgent ​

Answer 1

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\: \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

$\rm = \: \dfrac{1}{2}[x + y + z]\bigg[ {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}\bigg]$

$\large\underline{\sf{Solution-}}$

Consider RHS

$\rm \: \dfrac{1}{2}[x + y + z]\bigg[ {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}\bigg]$

$\rm = \: \dfrac{1}{2}[x + y + z]\bigg[ {x}^{2} + {y}^{2} - 2xy + {y}^{2} + {z}^{2} - 2yz + {z}^{2} + {x}^{2} - 2zx \bigg]$

$\rm = \: \dfrac{1}{2}[x + y + z]\bigg[ 2{x}^{2} + 2{y}^{2} + 2 {z}^{2} - 2xy - 2yz- 2zx\bigg]$

$\rm = \: \dfrac{1}{2} \times 2[x + y + z]\bigg[ {x}^{2} +{y}^{2} + {z}^{2} - xy - yz- zx\bigg]$

$\rm = \: [x + y + z]\bigg[ {x}^{2} +{y}^{2} + {z}^{2} - xy - yz- zx\bigg]$

$\rm \: = \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

Hence, Proved

Alternative Method

Consider LHS

$\rm :\longmapsto\: \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

$\rm = \: [x + y + z]\bigg[ {x}^{2} +{y}^{2} + {z}^{2} - xy - yz- zx\bigg]$

On multiply and divide by 2, we get

$\rm = \: \dfrac{1}{2} \times 2[x + y + z]\bigg[ {x}^{2} +{y}^{2} + {z}^{2} - xy - yz- zx\bigg]$

$\rm = \: \dfrac{1}{2}[x + y + z]\bigg[ 2{x}^{2} + 2{y}^{2} + 2 {z}^{2} - 2xy - 2yz- 2zx\bigg]$

$\rm = \: \dfrac{1}{2}[x + y + z]\bigg[ {x}^{2} + {x}^{2} + {y}^{2} + {y}^{2} + {z}^{2} + {z}^{2} - 2xy - 2yz- 2zx\bigg]$

$\rm = \: \dfrac{1}{2}[x + y + z]\bigg[({x}^{2} +{y}^{2} - 2xy) +({y}^{2} + {z}^{2} - 2yz )+ ({z}^{2} + {x}^{2} - 2zx)\bigg]$

$\rm = \: \dfrac{1}{2}[x + y + z]\bigg[ {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}\bigg]$

Hence, Proved

Additional Information :-

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)