Question

PLS GET THIS DONE!
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Answer 1

$\displaystyle\longrightarrow\sf{\dfrac{1}{2}(x+y+z)\left[(x-y)^2+(y-z)^2+(z-x)^2\right]}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{2}(x+y+z)[x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2]}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{2}(x+y+z)[2x^2+2y^2+2z^2-2xy-2yz-2zx]}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{2}\cdot2(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$

$\displaystyle\longrightarrow\sf{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$

$\displaystyle\longrightarrow\sf{\underline{\underline{x^3+y^3+z^3-3xyz}}}$

Answer 2

Step-by-step explanation:

Given :

• 1/2 ( x + y + z ) [ (x–y)² + (y–z)² + (z–x)² ]

Solutuon:

★ Using ( a – b )² = a² + b² + 2ab

→1/2 ( x + y + z ) ( x² + y² –2xy) + ( y² + z² + 2yz ) + ( z² + x² –2zx)

→ 1/2 ( x + y + z ) ( 2x² + 2y² + 2z² – 2xy – 2yz – 2zx )

→ 1/2 ( x + y + z ) 2 ( x² + y² + z² – xy – yz – zx )

→ (1/2)2 ( x + y + z ) ( x² + y² + z² – xy – yz – zx )

→ ( x + y + z ) ( x² + y² + z² – xy – yz – zx )

★ We know the formula that ★

$\small\implies{\sf }$ x³ + y³ + z³ – 3xyz = ( x + y + z ) ( x² + y² + z² – xy – yz – zx )

Therefore, The value of given equation will be x³ + y³ + z³ – 3xyz