Question

If x²+y²+z²=40 and xy+yz+zx=12,then find the value of x³+y³+z³-3xyz​

Answer 1

Given

⇒x² + y² + z² = 40

⇒xy + yz + zx = 12

To find

⇒The value of x³ + y³ + z³ - 3xyz

Identities we use

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

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

Using this identities , we can write as  

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

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

First of all we have to find the value of (x + y + z) , so we use

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

Now Put the value

⇒(x + y + z)² = 40 + 2(12)

⇒(x + y + z)² = 40 + 24

⇒(x + y + z)² = 64

⇒x + y + z = 8

Now  put the value on

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

⇒x³ + y³ + z³ - 3xyz​ = (8)(40 - { xy + yz + zx })

⇒x³ + y³ + z³ - 3xyz​ = 8(40 - 12)

⇒x³ + y³ + z³ - 3xyz​ = 8(28)

⇒x³ + y³ + z³ - 3xyz​ = 224

Answer

⇒x³ + y³ + z³ - 3xyz​ = 224

Answer 2

Step-by-step explanation:

ANSWER ✍️

Given

⇒x² + y² + z² = 40

⇒xy + yz + zx = 12

To find

⇒The value of x³ + y³ + z³ - 3xyz

Identities we use

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

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

Using this identities , we can write as  

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

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

First of all we have to find the value of (x + y + z) , so we use

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

Now Put the value

⇒(x + y + z)² = 40 + 2(12)

⇒(x + y + z)² = 40 + 24

⇒(x + y + z)² = 64

⇒x + y + z = 8

Now  put the value on

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

⇒x³ + y³ + z³ - 3xyz = (8)(40 - { xy + yz + zx })

⇒x³ + y³ + z³ - 3xyz = 8(40 - 12)

⇒x³ + y³ + z³ - 3xyz = 8(28)

⇒x³ + y³ + z³ - 3xyz = 224

Answer

⇒x³ + y³ + z³ - 3xyz = 224

hope this helps you!!

thank you ⭐