Question

If X+y+z=1,XY+yz+zx=-1 and xyz=-1,find the value of x³+y³+z³​

Answer 1

Step-by-step explanation:

x+y+z=1

x+y=-z

y+z=-x

z+x=-y

x³+y³+z³=(x+y+z)³-3(y+z)(z+x)(x+y)

=(1)³-3(-x)(-y)(-z)

=1+3xyz

=-2

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: