Question 196229
sol:
step1:
------
first take the x^2-x-6
 factors of x^2-x-6 are
x^2-3x+2x-6  (i.e; on multiplying we get -6 and adding we get -1)
x(x-3)+2(x-3)
(x+2)(x-3)

step2:
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take 4x^3-5x^2
in this x^2 is common term
this can be written as
x^2(4x-5)

step3:
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take 16x^3-25x
here x is common term
this can be written as
x(16x^2-25)
x[(4x)^2-(5)^2]
x[(4x+5)(4x-5)]  (i.e; a^2-b^2=(a+b)*(a-b))

step4:
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take 3x-9
here 3 is common term
this can be written as
3(x-3)
substitute all these values in the given expression
x^2-x-6                16x^3-25x
--------        *      -----------
4x^3-5x^2                  3x-9

(x+2)(x-3)           x(4x+5)(4x-5)
----------      *  ---------------

x^2(4x-5)              3(x-3)
here the  canciling terms are (x-3)(4x-5)andx
after canciling the remaining terms are
(x+2)*(4x+5)    4x*x+5*x+4x*2+5*2
------------ = ------------------
   3x                   3x
                         
4x^2+5x+8x+10
-------------
    3x       
4x^2+13x+10
-----------
    3x