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

[anemone]

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

 

[kaliprasad]

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

Spoiler

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)
= $x-xy-xz +xyz + y + z - yz$
= $x(1-y-z+yz) + (y+z-yz)$
=$x(1-y)(1-z) - (1-y)(1-z)+1$
= $1 + (x-1)(1-y)(1-z)$
= $1- (1-x)(1-y)(1-z)$

 

[anemone]

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

Spoiler

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!