1. If $$(1/x) + (1/y) + (1/z) = 0$$ and $$x + y + z = 11$$, then what is the value of $$x^{3}+y^{3}+z^{3}-3xyz$$ ?
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By: anil on 05 May 2019 02.26 am
Given : $$frac{1}{x}+frac{1}{y}+frac{1}{z}=0$$ => $$frac{yz+zx+xy}{xyz}=0$$ => $$xy+yz+zx=0$$ -------------(i) Also, $$x+y+z=11$$ ------------(ii) Squaring both sides, we get : => $$(x+y+z)^2=(11)^2$$ => $$(x^2+y^2+z^2)+2(xy+yz+zx)=121$$ Substituting value from equation (i), => $$x^2+y^2+z^2=121$$ ------------(iii) To find : $$x^{3}+y^{3}+z^{3}-3xyz$$ = $$(x+y+z)[(x^2+y^2+z^2)-(xy+yz+zx)]$$ Substituting values from equations (i), (ii) and (iii), = $$(11)(121-0)$$ = $$11 imes121=1331$$ => Ans - (A)
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