Question 708236
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I'll just do the first one.  Here is Pascal's triangle
written to the 6th line:
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            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 

(3a<sup>2</sup> + b)<sup>4</sup>
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The two terms are (3a<sup>2</sup>) 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:
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 (3a<sup>2</sup>) (b)  +  (3a<sup>2</sup>) (b)  +  (3a<sup>2</sup>) (b)  +  (3a<sup>2</sup>) (b)  +  (3a<sup>2</sup>) (b)
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Put in exponents from 4 down to 0 on the (3a²)'s:
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 (3a<sup>2</sup>)<sup>4</sup>(b)  +  (3a<sup>2</sup>)<sup>3</sup>(b)  +  (3a<sup>2</sup>)<sup>2</sup>(b)  +  (3a<sup>2</sup>)<sup>1</sup>(b)<sup>3</sup> + 1(3a<sup>2</sup>)<sup>0</sup>(b)<sup>4</sup>
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Put in exponents from 0 up to 4 on the (b)'s:
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 (3a<sup>2</sup>)<sup>4</sup>(b)<sup>0</sup> +  (3a<sup>2</sup>)<sup>3</sup>(b)<sup>1</sup> +  (3a<sup>2</sup>)<sup>2</sup>(b)<sup>2</sup> +  (3a<sup>2</sup>)<sup>1</sup>(b)<sup>3</sup> +  (3a<sup>2</sup>)<sup>0</sup>(b)<sup>4</sup>
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Put in the coefficients from the 4th line of Pascal's triangle, 1,4,6,4,1:
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1(3a²)<sup>4</sup>(b)<sup>0</sup> + 4(3a<sup>2</sup>)<sup>3</sup>(b)<sup>1</sup> + 6(3a<sup>2</sup>)<sup>2</sup>(b)<sup>2</sup> + 4(3a²)<sup>1</sup>(b)<sup>3</sup> + 1(3a<sup>2</sup>)<sup>0</sup>(b)<sup>4</sup>
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Simplify each term:
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1·3<sup>4</sup>a<sup>8</sup>·1    +  4·3<sup>3</sup>a<sup>6</sup>·b   +  6·3<sup>2</sup>a<sup>4</sup>·b<sup>2</sup>  +   4·3a<sup>2</sup>·b<sup>3</sup>  +   1·1·b<sup>4</sup>

   81a<sup>8</sup>     +  4·27a<sup>6</sup>·b   +   6·9a<sup>4</sup>·b<sup>2</sup>  +    12a<sup>2</sup>·b<sup>3</sup>  +      b<sup>4</sup>

   81a<sup>8</sup>     +   108a<sup>6</sup>b    +     54a<sup>4</sup>b<sup>2</sup>  +    12a<sup>2</sup>b<sup>3</sup>   +      b<sup>4</sup>  
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Edwin</pre>