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\((x+ay)(x+ay)=(x+ay)^2\)

 

And remember that:

\((a+b)^2 = a^2+2ab+b^2\)

 

So using this rule, we can just substitute a = x and ay = b

 

\(x^2+2xay+ay^2\)

 

You can use the rule to skip the process of using the distributive rule :)

Expand the following:
(x+a y) (x+a y)

 

(x+a y) (x+a y) = (x+a y)^2:
(x+a y)^2

 

(a y+x) (a y+x) = (a y) (a y) + (a y) (x) + (x) (a y) + (x) (x):
a y a y+a y x+x a y+x x

 

a y a y = a^2 y^2:
a^2 y^2+a y x+x a y+x x

 

x x = x^2:
Answer: |  a^2y^2 + 2ayx + x^2

\((x+ay)(x+ay)\\=x(x+ay)+ay(x+ay)\\=x^2+axy+axy+a^2y^2\\=x^2+2axy+a^2y^2\)

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