Question 86

If $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$ and x + y + z = 9, then what is the value of $$x^3 + y^3 + z^3 - 3xyz$$ ?

Solution

Given : $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$

=> $$\frac{yz+zx+xy}{xyz}=0$$

=> $$xy+yz+zx=0$$ -----------(i)

Also, $$x+y+z=9$$

Squaring both sides,

=> $$(x+y+z)^2=(9)^2$$

=> $$x^2+y^2+z^2+2(xy+yz+zx)=81$$

=> $$x^2+y^2+z^2+2(0)=81$$

=> $$x^2+y^2+z^2=81$$

To find : $$x^3 + y^3 + z^3 - 3xyz$$

= $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

= $$(9)(81-0)=729$$

=> Ans - (C)

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