SOLUTION: Pascals Triangle :Please i need help with this thanks solve: 1)(3a^2+b)^4 2.)(4x-3y)^3 3.)( square root 2 only then x+y)^6 (2x)^2 (5y)^3 +5(2x) (5y)^4 +5y)^5

I'll just do the first one.  Here is Pascal's triangle
written to the 6th line:

            1
          1   1
        1   2   1
      1   3   3   1
    1   4   6   4   1
  1   5  10  10   5    1`
1   6  15  20  15   6    1 

(3a2 + b)4

The two terms are (3a2) and (b). 
The power is 4 so we will have 1 more term than the power, so
write those terms side by side 5 times skipping a space between
and on the left and right for coefficients and exponents.  Put
+ signs between them:

 (3a2) (b)  +  (3a2) (b)  +  (3a2) (b)  +  (3a2) (b)  +  (3a2) (b)

Put in exponents from 4 down to 0 on the (3a²)'s:

 (3a2)4(b)  +  (3a2)3(b)  +  (3a2)2(b)  +  (3a2)1(b)3 + 1(3a2)0(b)4

Put in exponents from 0 up to 4 on the (b)'s:

 (3a2)4(b)0 +  (3a2)3(b)1 +  (3a2)2(b)2 +  (3a2)1(b)3 +  (3a2)0(b)4

Put in the coefficients from the 4th line of Pascal's triangle, 1,4,6,4,1:

1(3a²)4(b)0 + 4(3a2)3(b)1 + 6(3a2)2(b)2 + 4(3a²)1(b)3 + 1(3a2)0(b)4

Simplify each term:

1·34a8·1    +  4·33a6·b   +  6·32a4·b2  +   4·3a2·b3  +   1·1·b4

   81a8     +  4·27a6·b   +   6·9a4·b2  +    12a2·b3  +      b4

   81a8     +   108a6b    +     54a4b2  +    12a2b3   +      b4  

Edwin