Question

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

Answer 1

Answer:

$\large{\underline{\boxed{\sf x^3 + y^3 + z^3 = - 5}}}$

Step-by-step explanation:

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).

_______________________

★ Identity Used -

• a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - xy - yz - zx)
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

$\large{\underline{\underline{\mathfrak{\heartsuit \: Algebraic \: Identities : \: \heartsuit}}}}$

• (a - b)³ = a³ - b³ - 3ab (a - b)
• (a + b)³ = a³ + b³ + 3ab (a + b)
• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (x + a)(x + b) = x² + x(a + b) + ab

Answer 2

Answer:

- 5 .

Step-by-step explanation:

Given :

$\large x+y+z=1 \ ...(i)$

$\large xy+yz+zx=1 \ ..(ii)$

$\large xyz=-1 \ ..(iii)$

We have to find $\large x^3+y^3+z^3$

Using identity here

$\large a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ca)(a+b+c)+3abc\\\\\\\large \text{Putting value of ( i ) , ( ii ) and ( iii ) we get}\\\\\\\large x^3+y^3+z^3=(1)^3-3(1)(1)+3(-1)\\\\\\\large x^3+y^3+z^3=1-3-3\\\\\\\large x^3+y^3+z^3=1-6\\\\\\\large x^3+y^3+z^3=-5$

Thus we get answer - 5 .