$y^2 - 343 = x^{3} - 3(x^{2})(7) + 3(x)(49)-343$
to
$y^2 - 7^3 = (x-7)^3$

Is there a specific name for the second one? Could you please explain how to get from the first to the second.

2. Originally Posted by ericforman65
$y^2 - 343 = x^{3} - 3(x^{2})(7) + 3(x)(49)-343$
to
$y^2 - 7^3 = (x-7)^3$

Is there a specific name for the second one? Could you please explain how to get from the first to the second.
Hi ericforman65,

$y^2-343=y^2-7^3$ on the left.

The right side was obtained from $x^3-3x^3(7)+3x^2(49)+349$ by recognizing the binomial expansion of $(x-7)^3$

Remember the binomial theorem?

$(x-a)^3=x^3a^0-3x^2a^1+3x^1a^2-a^3$

Now, let a = 7

$(x-7)^3=x^3(7^0)-3x^2(7^1)+3x(7^2)-7^3$

$(x-7)^3=x^3-3x^2(7)+3x(49)-343$