Question

Hi GYES can any one tell me this solution x^3+y^3+z^3-3xyz if x+y+z=12 and x^2+y^2+z^2=70 please note spam it is 90 points answer ​

Answer 1

Answer:

396

Step-by-step explanation:

We have:

x + y + z = 12 and x² + y² + z² = 70.

Then:

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

⇒ (12)² = 70 + 2(xy + yz + zx)

⇒ 144 - 70 = 2(xy + yz + zx)

⇒ 74 = 2(xy + yz + zx)

⇒ xy + yz + zx = 37

Now,

x³ + y³ + z³ - 3xyz

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

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

= (12)(70 - 37)

= 12 * 33

= 396

Hope it helps!

Answer 2

x³ + y³ + z³ - 3xyz

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

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

= (12)(70 - 37)

= 12 * 33

= 396