If none of a,b,c,x,y,za,b,c,x,y,za,b,c,x,y,z is zero and x2(y+z)a3=y2(z+x)b3=z2(x+y)c3=xyzabc=1\frac { { x }^{ 2 }(y+z) }{ { a }^{ 3 } } =\frac { { y }^{ 2 }(z+x) }{ { b }^{ 3 } } =\frac { { z }^{ 2 }(x+y) }{ { c }^{ 3 } } =\frac { xyz }{ abc } =1a3x2(y+z)=b3y2(z+x)=c3z2(x+y)=abcxyz=1.
Prove that a3+b3+c3+abc=0{a}^3+{b}^3+{c}^3+abc=0a3+b3+c3+abc=0.
This a part of my set NMTC 2nd Level (Junior) held in 2014.
Note by Siddharth G 6 years, 7 months ago
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x2(y+z)=a3x^{2}(y+z)=a^{3}x2(y+z)=a3 and xyz=abcxyz=abcxyz=abc. Dividing the first equation by the second, we get x(1y+1z)=a2bcx (\frac {1}{y}+\frac {1}{z})=\frac {a^{2}}{bc}x(y1+z1)=bca2. We can do this two more times with different variables to get two more similar equations. Multiplying all of them,
xyz(1y+1z)(1y+1x)(1x+1z)=(x+y)(y+z)(x+z)xyz=1xyz (\frac {1}{y}+\frac {1}{z})(\frac {1}{y}+\frac {1}{x})(\frac {1}{x}+\frac {1}{z})=\frac {(x+y)(y+z)(x+z)}{xyz}=1xyz(y1+z1)(y1+x1)(x1+z1)=xyz(x+y)(y+z)(x+z)=1 hence
(x+y)(y+z)(x+z)=xyz(x+y)(y+z)(x+z)=xyz(x+y)(y+z)(x+z)=xyz. Expanding and subtracting both sides by xyzxyzxyz, then substituting xyzxyzxyz with abcabcabc, x2(y+z)x^{2}(y+z)x2(y+z) with a3a^{3}a3 and so on gives the desired result.
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x2(y+z)=a3 and xyz=abc. Dividing the first equation by the second, we get x(y1+z1)=bca2. We can do this two more times with different variables to get two more similar equations. Multiplying all of them,
xyz(y1+z1)(y1+x1)(x1+z1)=xyz(x+y)(y+z)(x+z)=1 hence
(x+y)(y+z)(x+z)=xyz. Expanding and subtracting both sides by xyz, then substituting xyz with abc, x2(y+z) with a3 and so on gives the desired result.