Question

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

Answer 1

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