Question

$\sf{Solve \: = \: \: \dfrac{{{\rm{m}} + {\rm{n}}}}{{{{\rm{m}}^2} + {\rm{mn\: }} + {{\rm{n}}^2}}} +\dfrac{{{\rm{m}} - {\rm{n}}}}{{{{\rm{m}}^2} - {\rm{mn\: }} + {{\rm{n}}^2}}}\ \textless \ br /\ \textgreater \ + {\rm{\: }}\dfrac{{2{{\rm{n}}^3}}}{{{{\rm{m}}^4} - {{\rm{m}}^2}{{\rm{n}}^2}{\rm{\: }} + {{\rm{n}}^4}}}}$

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Answer 1

No, (m−n)² actually means you’re squaring the whole dang thing! So let’s make x=m−n just to help you understand something. That means that we now have x² . That means that you have x⋅x .

Now, let’s do something for the next expression: m²−n². Let’s make x=m and y=n . So you’d then have x²−y², which equals x⋅x⋅y⋅y . That is obviously not the same as x⋅x .

If we actually simplify (m−n)² we get:

$\sf\red{(m−n)²=(m−n)⋅(m−n) }$

Now we use the FOIL method to simpify (First, Outer, Inner, Last). We multiply the first terms, then the outter, then the inner, and finally the last. We then add these all together.

$\sf\pink{m⋅m+(m⋅−n)+(m⋅−n)+n⋅n=m²−mn−mn+n2=m²−²mn+n²}$

That is obviously not the same as m²−n²

because it’s missing all its middle term!

Answer 2

Step-by-step explanation:

No, (m−n)² actually means you’re squaring the whole dang thing! So let’s make x=m−n just to help you understand something. That means that we now have x² . That means that you have x⋅x .

Now, let’s do something for the next expression: m²−n². Let’s make x=m and y=n . So you’d then have x²−y², which equals x⋅x⋅y⋅y . That is obviously not the same as x⋅x .

If we actually simplify (m−n)² we get:

\sf\red{(m−n)²=(m−n)⋅(m−n) }

Now we use the FOIL method to simpify (First, Outer, Inner, Last). We multiply the first terms, then the outter, then the inner, and finally the last. We then add these all together.

\sf\pink{m⋅m+(m⋅−n)+(m⋅−n)+n⋅n=m²−mn−mn+n2=m²−²mn+n²}

That is obviously not the same as m²−n²

because it’s missing all its middle term!