Question 279759
It is not true. So {{{A^2+B^2<>(A+B)^2}}} in general. To show that it is false, simply find values of A and B which contradict the equation.


By the way, the only solutions to {{{A^2+B^2=(A+B)^2}}} are when either {{{A=0}}} or {{{B=0}}}. If {{{A=0}}}, then {{{0^2+B^2=(0+B)^2}}} which then becomes {{{B^2=B^2}}} which is true for all B. This argument works similarly when {{{B=0}}}. For any other values of A and B, the equation is false.



Here's why the only solutions are {{{A=0}}} or {{{B=0}}}:



{{{A^2+B^2=(A+B)^2}}} Start with the given equation.



{{{A^2+B^2=A^2+2AB+B^2}}} FOIL



{{{0=2AB}}} Subtract {{{A^2}}} from both sides. Subtract {{{B^2}}} from both sides.



{{{2AB=0}}} Rearrange the equation.



{{{AB=0}}} Divide both sides by 2.



{{{A=0}}} or {{{B=0}}} Use the zero product property.