Question 1012113
Since a, b, and c are in an arithmetic progression, we can write them as
a, a + d, and a + 2d.
Since x, y, and z are in geometric progression , we can write them as
x, xr and xr^2.
Thus we can substitute into
x^b y^c z^a = x^c y^a z^b
and get
{{{x^(a+d)*(xr)^(a+2d)*(xr^2)^a = x^(a+2d)*(xr)^a*(xr^2)^(a+d)}}}
which yields
{{{x^(a+d)*x^(a+2d)*x^a*r^(a+2d)*r^(2a) = x^(a+2d)*x^a*x^(a+d)*r^a*r^(2a+2d)}}}
Now collect like bases and get
{{{x^(3a+3d)*r^(3a+2d) = x^(3a+3d)*r^(3a+2d)}}}
and we're done...