(5 b^{0} c^{-2})^{-3} (3 b^{3} a^{3})^{5} equals na^rb^sc^t wheres N AND THE EXPONENT OF a AND s the exponent of b and t the exponent of c

Question

Where n, the leading coefficient and r, the exponent of a, is and s, the exponent of b, is and finally t, the exponent of c, is

Answer 1

I hope I read the problem correctly:

 

[(5b^0c^-2)^-3][(3b^3a^3)^5]

 

First, we'll work on simplifying the two [ ] by using the 'power of a power' property of exponents which states (x^a)^b is x^(ab).

 

[5^-3b^0c^6][3^5b^15a^15]

 

Second, recall that (x^a)(x^b) = x^(a+b) and that x^0 =1

 

(5^-3)(3^5)b^15c^6a^15

 

Third, do the math on the numbers and recall that a negative exponent means you need to drop that term to the denominator.

 

(243/125)a^15b^15c^6.

 

So, n=(243/125), r=15, s=15, and t=6.