Prime factorization of a number

Hello. What have you tried or thought about with this exercise? The brute-force method is to evaluate the given expression, followed by factoring the result.

 
Hello. What have you tried or thought about with this exercise? The brute-force method is to evaluate the given expression, followed by factoring the result.

Hi, i tried the brute method :LOL:. So i was thinking what if the expression inside the parentheses is like REALLY big then what? There must a faster way to solve this. That's why i posted it here. I got the right answer which is option (b) after evaluating the expression.
 
Hi, I need help with this question.

View attachment 32893
Start by factoring [imath]3^2[/imath] out of [imath]3^4+3^2[/imath], and writing this number in factored form (or just evaluate it and factor, since it isn't big).

Then you just have a product of factors, and can apply rules for exponents. Give it a try, and show us an attempt, so we can help more.
 
3^4 + 3^2 = 81 + 9 = 90 = 9*10 = 3^2*2*5. So what is the prime factorization of (3^4 + 3^2)^3.
Hint: It is the prime factorization of (3^2*2*5)^3
 
Hi, i tried the brute method :LOL:. So i was thinking what if the expression inside the parentheses is like REALLY big then what? There must a faster way to solve this. That's why i posted it here. I got the right answer which is option (b) after evaluating the expression.
Brute force is in the eyes of the beholder. Here is my solution, with no large numbers
[imath](3^4+3^2)^3=(3^6)(3^2+1)^3=(3^6)(10)^3=2^3 \cdot 3^6\cdot 5^3 [/imath]
Thus [imath]{\bf\large(3^4+3^2)^3\cdot 5^2=2^3 \cdot 3^6\cdot 5^5} [/imath]

[imath][/imath][imath][/imath][imath][/imath][imath][/imath]
 
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