Question 1105288
{{{x^2-6x+1=0}}} can be solved by "completing the square"
or by using the quadratic formula.
Completing the square:
{{{x^2-6x+1=0}}}
{{{x^2-6x=-1}}}
{{{x^2-6x+9=-1+9}}}
{{{(x-3)^2=8}}}
{{{x-3=" " +- sqrt(8)}}}
{{{x=3 +- sqrt(8)}}}
The expression {{{log(a,(ab^2))+log(b,(ba^2))}}} becomes itself if we interchange a  and b,
so it does not matter which root is {{{log((a))}}} and which root is {{{log((b))}}} .
Let us say that {{{log((a))=3+sqrt(8)}}} and {{{log((b))=3-sqrt(8)}}} . 
Those are base 10 logarithms, so to get to base a and base b,
we use the "change of base formula."
{{{log(b,a)}}}{{{"="}}}{{{log((a))/log((b))}}}{{{"="}}}{{{(3+sqrt(8))/(3-sqrt(8))=(3+sqrt(8))^2/((3+sqrt(8))(3-sqrt(8)))=(9+8+6sqrt(8))/(9-8)=17+6sqrt(8)}}}
{{{log(a,b)}}}{{{"="}}}{{{log((b))/log((a))}}}{{{"="}}}{{{(3-sqrt(8))/(3+sqrt(8))=(3-sqrt(8))^2/((3+sqrt(8))(3-sqrt(8)))=(9+8-6sqrt(8))/(9-8)=17-6sqrt(8)}}} .
 
Now we can calculate that expression
 {{{log(a,(ab^2))+log(b,(ba^2))}}}{{{"="}}}{{{log(a,a)+2*log(a,b)+log(b,b)+2*log(b,a)}}}{{{"="}}}{{{1+2*log(a,b)+1+2*log(b,a)}}}{{{"="}}}{{{2+2(log(a,b)+log(b,a))}}}{{{"="}}}{{{2(1+log(a,b)+log(b,a))}}}{{{"="}}}{{{2(1+17+6sqrt(8)+17-6sqrt(8))}}}{{{"="}}}{{{2(1+17+17)}}}{{{"="}}}{{{2*35}}}{{{"="}}}{{{highlight(70)}}}