How do I simplify in the binomial therom

Question

x2 + 3)6 = 6C0 (x2)6(3)0 + 6C1 (x2)5(3)1 + 6C2 (x2)4(3)2 + 6C3 (x2)3(3)3

+ 6C4 (x2)2(3)4 + 6C5 (x2)1(3)5 + 6C6 (x2)0(3)6

   

Answer 1

Not sure what part you're having trouble simplifying, but I'll show you how to do the 6C3(x²)³(3)³ term  

 

6C3 = 6!/(3!(6-3)!) = 6*5*4/(3*2*1) = 20

(x²)³ = x⁶  

(3)³ = 27  

 

so 6C3(x²)³(3)³ = 20*x⁶*27 = 540x⁶  

 

You can simplify the other terms in a similar way