Question

if x+y+z=1,xy+yz+zx=-1 and xyz=-1.Find the value of x³+y³+z³

Answer 1

first ,we find x^2+y^2 +z^2
(x+y+z)^2=x^2 +y^2+z^2+2xy +2yz +2zx
(1)^2         =x^2+y^2+z^2+2(1)
1              =  x^2+y^2+z^2+2
1-2            =  x^2+y^2+z^2
-1              =  x^2+y^2+z^2

x^3+ y^3 +z^3-3xyz=(x+y+z)(x^2+y^2 +z^2-xy-yz-zx)
1^3-3*1=(1)(-1-1)
1-3=-2
-2=-2
-2+2=0
0 is the answer

Answer 2

Answer:

Given that :

  x + y + z = 1

  xy + yz + zx = 1

  xyz = (-1)

We know that ;

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

Here, we need to find (x² + y² + z²) first,

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

⇒ (1)² = x² + y² + z² + 2(1)

⇒ 1 = x² + y² + z² + 2

⇒ x² + y² + z² = 1 - 2

⇒ x² + y² + z² = (-1)

Now,

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

⇒ x³ + y³ + z³ - 3(-1) = 1 {(-1) - (1)}

⇒ x³ + y³ + z³ + 3 = 1 * (-2)

⇒ x³ + y³ + z³ + 3 = - 2

⇒ x³ + y³ + z³ = - 2 - 3

⇒ x³ + y³ + z³ = -5

Hence, the answer is (-5).

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Step-by-step explanation: