Question

$x { }^{3} + 4 { }^{3} + 2 {}^{3 } - 3 \times \: yz = 1 \2(x + y + z + )((x - y {}^{2} ) + (y - z) {}^{2} + (z - x {}^{2} ))$
please help me ❤️​

Answer 1

Correct Question :-

x³ + y³+ z³ - 3xyz = $\dfrac{1}{2}$ (x + y+ z) [(x - y)² + (y - z)² + (z - x)³]

*To prove this identity, we need to take help of another identity*

We know that:-

x³ + y³ + z³- 3xyz

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

Now, we just need to change (x² + y² + z² - xy - yz - zx) as the sum of square term.

By solving further :-

x² + y² + z² - xy - yz - zx

= 1/2 (2x² + 2y² + 2z² - 2xy - 2yz - 2zx)

= 1/2 (x - 2 x y +y² +y² - 2yz + z² + z² - 2zx + x)

= 1/2 [(x - y)² + (y - z)² + (z - x)³]

So we get:-

x³ + y³+ z³ - 3xyz = $\dfrac{1}{2}$ (x + y+ z) [(x - y)² + (y - z)² + (z - x)³]

*Hence,Proved*

Answer 2

Answer:

x³ + y³+ z³ - 3xyz = \dfrac{1}{2}

2

1

(x + y+ z) [(x - y)² + (y - z)² + (z - x)³]

*To prove this identity, we need to take help of another identity*

We know that:-

x³ + y³ + z³- 3xyz

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

Now, we just need to change (x² + y² + z² - xy - yz - zx) as the sum of square term.

By solving further :-

x² + y² + z² - xy - yz - zx

= 1/2 (2x² + 2y² + 2z² - 2xy - 2yz - 2zx)

= 1/2 (x - 2 x y +y² +y² - 2yz + z² + z² - 2zx + x)

= 1/2 [(x - y)² + (y - z)² + (z - x)³]

So we get:-

x³ + y³+ z³ - 3xyz = \dfrac{1}{2}

2

1

(x + y+ z) [(x - y)² + (y - z)² + (z - x)³]

*Hence,Proved*