<PRE><B>Question 63538</B>
1.) 5x^3/125x^3 
2.) 343t^9/21t^2

Consider the first one.

1.) 5x^3/125x^3  = 5 * x * x * x / 5 * 5 * 5 *x* x* *x 

I have expanded both the numerator and the denominator, we get

we are left with after cancecellation,

1 / 5 * 5 = 1 / 25

Hence 5x^3/125x^3 = 1 / 25


2) 343t^9/21t^2 = 7 * 7 *7 * t* t* t*t*t*t*t*t*t / 7 * 3 t * t

                = 7 * 7 * t* t* t*t*t*t*t / 3

                = 49 t^7 / 3      

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