Question

If x+y+z=3 and x² + y² + z² 101
then what is the value of square root of (x² + y + z² – 3xyz )​

Answer 1

Step-by-step explanation:

Given If x+y+z=3 and x² + y² + z² = 101

then what is the value of square root of (x^3 + y^3 + z^3 – 3xyz )​

• We know certain formula for x^3 + y^3 + z^3
• We can write this as x^3 + y^3 + z^3 – 3xyz = (x + y + z)[(x^2 + y^2 + z^2 – xy – yz – za)
• Using the formula (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
• Substituting the values we get (3)^2 = 101 + 2(xy + yz + zx)
• Or 2(xy + yz + zx) = 9 – 101 = - 92 / 2
•                                               = - 46
• So we know the formula for  
• Or x^3 + y^3 + z^3 – 3abc = (x + y + z)[(x + y + z)^2 – 3 (xy + yz + zx)]
•                                          = (3) [(3)^2 – 3 (- 46)]
•                                        = 3 (9 + 138)
•                                        = 3 (147)
•                                          = 441
• Or √x^3 + y^3 + z^3 – 3abc = √441
•                                             = 21

Reference link will be

https://brainly.in/question/11413505

Answer 2

√(x²+y²+z²-3xyz) = 21

• We know the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
• It's given = x+y+z = 3 and x²+y²+z²=101
• Putting the values, we get = 3² = 101 + 2(xy+yz+zx)
• xy + yz + zx = -46      (1)
• Now, the identity x² + y² + z² - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
• 3(101 - (-46)) = 147 × 3 = 441          (from 1)
• Therefore, √x² + y² + z² - 3xyz = √441 =21