Question 175710
{{{((8b^6)^(4/3)) / ((64a^12)^(1/6))}}}
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First you need the general concept
{{{(a^5)^3 = a^5 * a^5 * a^5}}}
{{{a^5 * a^5 * a^5 = a^15}}}
So, when you see something like this, you multiply
the exponents
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{{{((8b^6)^(4/3)) / ((64a^12)^(1/6))}}}
next
{{{(8b)^(6*(4/3)) / (64a)^(12*(1/6))}}}
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Now you need the general concept
{{{(a*b)^5 = (a*b)*(a*b)*(a*b)*(a*b)*(a*b)}}}
next
{{{(a*b)*(a*b)*(a*b)*(a*b)*(a*b) = a^5*b^5}}}
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{{{(8b)^(6*(4/3)) / (64a)^(12*(1/6))}}}
next
{{{(8b)^(8) / (64a)^(2)}}}
next
{{{(8^8)*(b^8) / ((64^2)*(a^2))}}}
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Now you need to go back to the 1st concept to deal
with {{{64^2}}} (only working backwards)
{{{64^2 = (8^2)^2}}}
next
{{{(8^2)^2 = 8^(2*2)}}}
next
{{{8^(2*2) = 8^4}}}
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{{{(8^8)*(b^8) / (64^2)*(a^2)}}}
next
{{{(8^8)*(b^8) / (8^4)*(a^2)}}}
This is the same as:
{{{((8^8)/(8^4)) * ((b^8)/ (a^2))}}}
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Another general concept:
{{{a^5 / a^3 = a^(5-3)}}}
next
{{{a^(5-3) = a^2}}}
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{{{((8^8)/(8^4)) * ((b^8)/ (a^2))}}}
next
{{{(8^4) * ((b^8)/ (a^2))}}}
I'm not sure if this is exactly
what they want, but the method should be OK