Question 1198057
<pre>
If you mean negative signs by those underscores, then what you want to
simplify is this:
 
{{{((a^2*b^(-3)*c)^"3/4")/(a^(-2)*b^3*c^5)^""}}}{{{""=""}}}{{{a^p*b^q*c^r}}}

However with 6 unknowns, we cannot be sure there is any way to determine 
p, q and r, so that there will be just one answer for p+2q-r.

Fraction exponents are a bummer, so let's clear the fraction exponent by 
raising both sides to the 4 power:

{{{(((a^2*b^(-3)*c)^"3/4")/(a^(-2)*b^3*c^5)^"")^4}}}{{{""=""}}}{{{(a^p*b^q*c^r)^4}}}

{{{((a^2*b^(-3)*c)^3)/(a^(-2)*b^3*c^5)^4=a^(4p)*b^(4q)*c^(4r)}}}

{{{(a^6*b^(-9)*c^3)/(a^(-8)*b^12*c^20)=a^(4p)*b^(4q)*c^(4r)}}}

Subtract exponents of like bases to divide on the left side:

{{{a^(6-(-8))b^(-9-12)c^(3-20)=a^(4p)b^(4q)c^(4r)}}}

{{{a^(6+8)b^(-21)c^(-17)=a^(4p)b^(4q)c^(4r)}}}

{{{a^14b^(-21)c^(-17)=a^(4p)b^(4q)c^(4r)}}}

Divide both side by the left side to get 1 on the left

{{{(a^14b^(-21)c^(-17))/(a^14b^(-21)c^(-17))=(a^(4p)b^(4q)c^(4r))/(a^14b^(-21)c^(-17))}}}

{{{1=(a^(4p)b^(4q)c^(4r))/(a^14b^(-21)c^(-17))}}}

Subtract exponents of like bases to divide on the right side:

{{{1=a^(4p-14)b^(4q-(-21))c^(4r-(-17))}}}

{{{1=a^(4p-14)b^(4q+21)c^(4r+17)}}}

Maybe your teacher thought you could set all those exponents = 0,
so we would have unique solutions for p,q,r, so we could find
the value of p+2q-r to be -11/4.  But that is only one solution.

For a = 1, b = 2, c = 3, p = 4, q = 5, then r≈-10.71702997, 
then p+2q-r = 24.71702997 approximately.

But if a = 2, b = 3, c = 4, p = 5, q = 6, r≈-10.71702997, 
then p+2q-r = -13.91541407 approximately.

There are infinitely many values possible for p+2q-r.

You should remind your teacher that we can't just assume that the
exponents of a, b, and c are equal to 0, just because the left side
is 1.  a, b, and c would have to be given.

Edwin</pre>