Question

If x³ + y³ + z³ = 3xyz and x + y + z = 0;

find the values of

(x + y)²/xy + ( y + z)²/yz + (z + x)²/zx

Answer 1

Answer:

Your question needs some correction.

Prove that,

x³ + y³ + z³ - 3xyz

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

Proof.

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) ...(i)

Now, we just need to change

(x² + y² + z² - xy - yz - zx)

as the sum of square term.

So, x² + y² + z² - xy - yz - zx

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

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

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

From (i), we get

x³ + y³ + z³ - 3xyz

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

Thus, confirmed.

I hope it helps you.

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Answer 2

Answer:

P^3 + q^3+r^3–3pqr

=( p+q+r)(1/2((p-q)^2+(q-r)^2+(r-p)^2))

=(x+y+z)(1/2*4*((x-y)^2 +(y-z^2+(z-x)^2)

= 4(x^3+y^3+z^3–3xyz)

=4(3(1+xyz)-3xyz)

=12