InterviewSolution
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InterviewSolution
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If X+y+z= 9, xy+yz+zx=26, find the value of x^3+y^3+z^3-3xyz |
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Answer» (X + Y + Z)^2= x^2+y^2+z^2+2(xy+yz+zx) 9^2. =x^2+y^2+z^2+2×26 81. = x^2+y^2+z^2+5281-52 = x^2+y^2+z^229. = x^2+y^2+z^2x^3+y^3+z^3-3xyz= (x+y+z)(x^2+y^2+z^2-(xy+yz+zx) =9(29-26) =9×3 x^3+y^3+z^3-3xyz= 27 ( X + Y + Z)^2= x^2+y^2+z^2+2( ( |
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