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Factor completely and write the answer with no negative exponents. Do not rationalize the denominator.



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Cancelling the common factor between the numerator and the denominator, the given expression, 6(t1)5(2t+5)66(2t+5)5(2)(t1)6[(2t+5)6]2,\dfrac{6(t-1)^5(2t+5)^6-6(2t+5)^5(2)(t-1)^6}{[(2t+5)^6]^2} , simplifies to

6(t1)5(2t+5)66(2t+5)5(2)(t1)6(2t+5)6(2)(use (am)n=amn)=6(t1)5(2t+5)66(2t+5)5(2)(t1)6(2t+5)12=(2t+5)5[6(t1)5(2t+5)6(2)(t1)6](2t+5)12(factor (2t+5)5)=(2t+5)5[6(t1)5(2t+5)6(2)(t1)6](2t+5)127(factor (2t+5)5)=6(t1)5(2t+5)6(2)(t1)6(2t+5)7=1(2t+5)76(t1)5(2t+5)6(2)(t1)6.\begin{align*} & \dfrac{6(t-1)^5(2t+5)^6-6(2t+5)^5(2)(t-1)^6}{(2t+5)^{6(2)}} &\left( \text{use }(a^m)^n=a^{mn} \right) \\\\&= \dfrac{6(t-1)^5(2t+5)^6-6(2t+5)^5(2)(t-1)^6}{(2t+5)^{12}} \\\\&= \dfrac{(2t+5)^5\left[ 6(t-1)^5(2t+5)-6(2)(t-1)^6 \right]}{(2t+5)^{12}} &\left( \text{factor }(2t+5)^5 \right) \\\\&= \dfrac{\cancel{(2t+5)^5}\left[ 6(t-1)^5(2t+5)-6(2)(t-1)^6 \right]}{(2t+5)^{\cancel{12}7}} &\left( \text{factor }(2t+5)^5 \right) \\\\&= \dfrac{ 6(t-1)^5(2t+5)-6(2)(t-1)^6 }{(2t+5)^{7}} \\\\&= \dfrac{ 1 }{(2t+5)^{7}}\cdot6(t-1)^5(2t+5)-6(2)(t-1)^6 .\end{align*}

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