MHB Prove Algebra Challenge: $(x,y,z,a,b,c)$ Equation

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The discussion revolves around proving the equation $x^3a + y^3b + z^3c = 1 - (1 - x)(1 - y)(1 - z)$ under the conditions that $a + b + c = ax + by + cz = x^2a + y^2b + z^2c = 1$. Participants clarify the correct formulation of the equation, emphasizing the need for accurate variable representation. The consensus is that the proof hinges on the established relationships among the variables. The challenge invites further exploration of algebraic identities and relationships. This mathematical inquiry highlights the interconnectedness of the variables in the given constraints.
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For reals $x,\,y,\,z$ and $a,\,b$ and $c$ that satisfy $a + b + c = ax + by + cz = x^2a + y^2b + z^2c = 1$,

prove that $x^3a + y^3b + cz^3c = 1 − (1 − x)(1 − y)(1 − z)$
 
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anemone said:
For reals $x,\,y,\,z$ and $a,\,b$ and $c$ that satisfy $a + b + c = ax + by + cz = x^2a + y^2b + z^2c = 1$,

prove that $x^3a + y^3b + cz^3c = 1 − (1 − x)(1 − y)(1 − z)$

I think you mean
prove that $x^3a + y^3b + z^3c = 1 − (1 − x)(1 − y)(1 − z)$

we are given
$a+b+c=\cdots(1)$
$ax+by+cz=1\cdots(2)$
$ax^2+by^2+cz^2 = 1\cdots(3)$
so we have
$x^3a+y^3b+z^3c$
= $x( 1- y^2b-z^2c) + y (1-x^2a-z^2c) + z(1-x^2a-y^2b)$ using (3)
= $x+y+z- xy(by+ax) - zx(cz+ax) - yz(by+cz)$
= $x+y+z-xy(1-cz) - zx(1-by) - yz(1-ax)$ using (2) in each of 3 expressions
= $x+y+z - xy - zx - yz + xyz(c+b+a)$
= $x+y+z - xy - zx - yz + xyz$ using (1)
= $x-xy-xz +xyz + y + z - yz$
= $x(1-y-z+yz) + (y+z-yz)$
=$x(1-y)(1-z) - (1-y)(1-z)+1$
= $1 + (x-1)(1-y)(1-z)$
= $1- (1-x)(1-y)(1-z)$
 
Last edited:
kaliprasad said:
I think you mean
prove that $x^3a + y^3b + z^3c = 1 − (1 − x)(1 − y)(1 − z)$

we are given
$a+b+c=\cdots(1)$
$ax+by+cz=1\cdots(2)$
$ax^2+by^2+cz^2 = 1\cdots(3)$
so we have
$x^3a+y^3b+z^3c$
= $x( 1- y^2b-z^2c) + y (1-x^2a-z^2c) + z(1-x^2a-y^2b)$ using (3)
= $x+y+z- xy(by+ax) - zx(cz+ax) - yz(by+cz)$
= $x+y+z-xy(1-cz) - zx(1-by) - yz(1-ax)$ using (2) in each of 3 expressions
= $x+y+z - xy - zx - yz + xyz(c+b+a)$
= $x+y+z - xy - zx - yz + xyz$ using (1)
= $x-xy-xz +xyz + y + z - yz$
= $x(1-y-z+yz) + (y+z-yz)$
=$x(1-y)(1-z) - (1-y)(1-z)+1$
= $1 + (x-1)(1-y)(1-z)$
= $1- (1-x)(1-y)(1-z)$

Perfect, kaliprasad!:cool:
 
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