1. If $$X+Y+Z=6$$ and $$XY+ZX+ZY=10$$, then find the value of $$X^3+Y^3+Z^3-3XYZ$$.
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By: anil on 05 May 2019 02.04 am
Given : $$xy+yz+zx=10$$ -------------(i) Also, $$x+y+z=6$$ ------------(ii) Squaring both sides, we get : => $$(x+y+z)^2=(6)^2$$ => $$(x^2+y^2+z^2)+2(xy+yz+zx)=36$$ Substituting value from equation (i), => $$x^2+y^2+z^2+2(10)=36$$ => $$x^2+y^2+z^2=36-20=16$$ ------------(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), = $$(6)(16-10)$$ = $$6 imes6=36$$ => Ans - (C)
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