Question

find the value of xcube +ycube+ zcube if x+y+z =1 and xy+yz+zx = -1 and xyz= -1​

Answer 1

Answer:

⇒ x³ + y³ + z³ = 1

Step-by-step explanation:

Given :

To find the value of :

x³ + y³ + z³.

If, x + y + z = 1,.

xy + yz + zx = -1,

xyz = -1.

Solution :

We know that,

(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc,.

Substituting,

a = x , b = y , c = z,.

We get,

(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc

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

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

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

⇒ (1)² - 2(-1) = x² + y² + z²

⇒ 1 + 2 = x² + y² + z²

⇒ 3 = x² + y² + z² ...(i)

We know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ac)

Substituting,

a = x , b = y , c = z,.

We get,

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

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

⇒ x³ + y³ + z³ = (1)[(3) - (-1)] + 3(-1)

⇒ x³ + y³ + z³ = (1)[3+1] - 3

⇒ x³ + y³ + z³ = (1)[4] - 3 = 4 - 3 = 1

⇒ x³ + y³ + z³ = 1