Question 283051
{{{((a+b)^2-c^2)/(3a+3b-3c)}}} Start with the given expression.



{{{((a+b+c)(a+b-c))/(3a+3b-3c)}}} Factor the numerator (using the difference of squares formula).



{{{((a+b+c)(a+b-c))/(3(a+b-c))}}} Factor out the GCF 3 from the denominator.



{{{((a+b+c)*highlight((a+b-c)))/(3*highlight((a+b-c)))}}} Highlight the common terms.



{{{((a+b+c)*cross((a+b-c)))/(3*cross((a+b-c)))}}} Cancel out the common terms.



{{{(a+b+c)/3}}} Simplify.



So {{{((a+b)^2-c^2)/(3a+3b-3c)}}} simplifies to {{{(a+b+c)/3}}}



In other words, {{{((a+b)^2-c^2)/(3a+3b-3c)=(a+b+c)/3}}}