Question

if X + Y + Z is equals to 9 and xy + Y Z plus ZX is equals to 23 then find the value of x cube + y cube + Z cube minus 3 x y z​

Answer 1

Answer:

HELLO THERE!!

x+y+z=9

(x+y+z)^2=81

x^2+y^2+z^2+2(xy+yz+zx)=81

x^2+y^2+z^2+2 (23)=81

=81-46

=35

x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx)

=(9)(35-23)

=9 (12)

=108

HOPE IT HELPS!! PLEASE MARK AS BRAINLIEST!!

Answer 2

Step-by-step explanation:

Given -

• x + y + z = 9
• xy + yz + zx = 23

→ -1(-xy - yz - zx) = 23

→ -xy - yz - zx = -23

To Find -

• Value of x³ + y³ + z³ - 3xyz

As we know that :-

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

→ (9)² = x² + y² + z² + 2(xy + yz + zx)

→ 81 = x² + y² + z² + 2(23)

→ 81 - 46 = x² + y² + z²

→ x² + y² + z² = 35

Now,

As we know that :-

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

→ (9)(35 - 23)

→ 9×12

→ 108

Hence,

The value of x³ + y³ + z³ - 3xyz is 108