Question 55

If x + y + z = 3, and $$x^2 + y^2 + z^2 = 101$$, then what is the value of $$\sqrt{x^3 + y^3 + z^3 - 3xyz}$$?

Solution

x³ + y³ + z³ - 3xyz = (x+y+z) [(x + y + z)² - 3(xy + yz + zx)] ---(1)

And,

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

On put the values,

$$3^2 = 101 + 2(xy + yz + zx)$$

(xy + yz + zx) = -92/2 = -46

From the eq(1),

x³ + y³ + z³ - 3xyz = (3) [(3)² - 3(-46)] = 3 $$\times 147 = 441

$$\sqrt{x^3 + y^3 + z^3 - 3xyz} = \sqrt{441}$$ = 21

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