Question 22

If $$x + y + z = 1, x^2 + y^2 + z^2 = 2$$ and $$x^3 + y^3 + z^3 = 3$$,then what is the value of $$xyz$$?

Solution

$$\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\ .$$

Or, $$\left(1\right)^2=2+2\left(xy+yz+xz\right).$$

Or, $$\left(xy+yz+xz\right)=-\frac{1}{2}.$$

$$x^3+y^3+z^3=(x+y+z)[x^2+y^2+z^2−xy−yz−xz]+3xyz\ .$$

Or, $$3=(1)[2+\frac{1}{2}]+3xyz\ .$$

Or, $$3xyz=3-\frac{5}{2}=\frac{1}{2}\ .$$

Or, $$xyz=\frac{1}{6}\ .$$

B is correct choice.

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