Can you help me out with this math problem? Show that [x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]=1?

please do not solve this using logarithm method; my teacher told me specifically only to solve this arithmetically

Can you help me out with this math problem? Show that [x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]=1?
x^a(b-c) = x^(ab - ac) = x^(ab)/x^(ac)





x^b(a-c) = x^(ab - bc) = x^(ab)/x^(bc)





Now, [x^a(b-c)]/[x^b(a-c)] =


= [x^(ab)/x^(ac)] / [x^(ab)/x^(bc)] =


= [1/x^(ac)]/[1/x^(bc)] =


= 1/x^(ac) * x^(bc) =


= x^(bc)/x^(ac) =


= [(x^b)^c]/[(x^a)^c] =


= [(x^b/x^a)^c]





QED
Reply:x^[a(b-c)]/[x^b(a-c)]=x^[ab-ac-ab+bc]=x^...


=[x^b/x^a)^c].


this easily leads to the final result desired.
Reply:[x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]


[x^ab-ac /x^ab-bc ] divided by [(x^b-a)^c]


x^ab-ac -ab +bc divided by x^c(b-a)


x^c(b-a) divided by x^c(b-a)


=1
Reply:Let's start with the first half of the problem:


[x^a(b-c)]/[x^b(a-c)]





You can bring the denominator to the top by making the exponent negative:


[x^a(b-c)] * {x^-[b(a-c)]}





When you do all the distributions, you get:


(x^ab-ac) * (x^-ba+bc) or


(x^ab-ac) * (x^-ab+bc)





When you multiply two (same) bases that have exponents, you simply add the exponents, keeping the base:


x^bc-ac





(The ab terms got canceled)





Now we can apply the same rules to the second part of the problem:


[(x^b/x^a)^c]





Remember that this whole term was in the denominator, so we can bring c to the top by reversing its sign:


[(x^b/x^a)^-c]





And we can do the same for a:


[(x^b * x^-a)^-c]=





Now when you have a power raised to another power, you simply multiply the exponents:


(x^-bc * x^ac)





Adding these exponents, we get:


(x^-bc+ac)





Now our full problem looks like:


x^(bc-ac) * x^(-bc+ac) = 1





Now when you add those exponents, you get zero:


x^0 = 1





And anything raised to the zero power is 1:


1 = 1