Question 696849
multiply and put in simplest form. 
1.) (b+4)/(2)^2 times (3)/(b+4)^3 
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[(b+4)/4] * [3/[(b+4)(b^2-4b+16)]]
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Cancel any factor(s) common to a numerator and a denominator:
[1/4] * [3/[(b^2-4b+16)]]
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= 3/[4(b^2-4b+16)]
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2.) (b+4)/(b-2) times (b^2)-(4)/(b^2)-(16)
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[(b+4)/(b-2)] * [[(b-2)(b+2)/(b-4)(b+4)]
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Cancel where you can:
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[1/1] * [[(b+2)/(b-4)]
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= (b+2)/(b-4)
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Cheers,
Stan H.
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