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(8x-8)(8x+8)
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\(64x^2 - 64 \)
remember this formula \(\Huge \color{pink}{(a+b)(a-b)=a^2-b^2}\) but if u wanna to find zero of \(\Huge \color{pink}{(y+b)(y-b)=0}\) then \(\Huge \color{pink}{ y=-b , y=b} \)
There are a few ways to proceed. Here's one that has not yet been suggested. (8x+8)(8x-8) = 8(x+1)8(x-1) = 8*8*(x+1)*(x-1) = 64(x^2 - 1)
@LarsEighner is correct
:)
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@cmckie Actually, ikram002p and tkhunny are more correct, since their responses were not in direct violation of he Code of Conduct.
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