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Helga [31]
4 years ago
7

Factor this expression completely, then place the factors in the proper location on the grid. a^3y + 1

Mathematics
1 answer:
ivolga24 [154]4 years ago
7 0

Answer:

a^{3y} + 1  = (a^{y}+1 )^{3}  - 3a^y(a^{y}+1)\\\\

Step-by-step explanation:

We are to factorize the expression a^{3y} + 1 completely. To do this, we will apply the expression below;

The expression can be rewritten as a^{3y} + 1^{3}

To factorize the expression, we need to first factorize (a^{y}+1 )^{3} first

(a^{y}+1 )^{3} =(a^{y}+1 )(a^{y}+1 )^{2}\\= (a^{y}+1 )((a^y)^{2}  } + 2a^{y} +1)\\= (a^y)^{3} +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y+1\\(a^{y}+1 )^{3}  = ((a^y)^{3} + 1) +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y\\(a^{y}+1 )^{3}  = ((a^y)^{3} + 1) +3(a^y)^{2}+3a^y\\

The we will make a^{3y} + 1^{3} the subject of the formula as shown;

(a^y)^{3} + 1 = (a^{y}+1 )^{3}  - (3(a^y)^{2}+3a^y)\\(a^y)^{3} + 1^{3}  = (a^{y}+1 )^{3}  - (3(a^y)^{2}+3a^y)\\\\

(a^y)^{3} + 1  = (a^{y}+1 )^{3}  - (3(a^y)^{2}+3a^y)\\\\

a^{3y} + 1  = (a^{y}+1 )^{3}  - (3(a^y)^{2}+3a^y)\\\\

a^{3y} + 1  = (a^{y}+1 )^{3}  - 3a^y(a^{y}+1)\\\\

This last result gives the expansion of the expression

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