User:Twentysand

A common mistake when learning about exponents:
$(a^{2}+b^{2})^{3}\neq a^{6}+b^{6}$

To see why, lets expand the outer exponent:
$(a^{2}+b^{2})^{3}=(a^{2}+b^{2})\times (a^{2}+b^{2})\times (a^{2}+b^{2})$

Actually lets do this - since we know we can use the FOIL method to Distribute a square of sums:
$(a^{2}+b^{2})^{3}=(a^{2}+b^{2})^{2}\times (a^{2}+b^{2})$

We can now FOIL the first part:
$(a^{2}+b^{2})^{2}=a^{4}+(a^{2})(b^{2})+(a^{2})(b^{2})+b^{4}$

So now we have $(a^{2}+b^{2})^{3}=(a^{4}+(a^{2})(b^{2})+(a^{2})(b^{2})+b^{4})\times (a^{2}+b^{2})$

$=(a^{4}+2a^{2}b^{2}+b^{4})\times (a^{2}+b^{2})$
$=(a^{6}+2a^{4}b^{2}+a^{2}b^{4}+a^{4}b^{2}+2a^{2}b^{4}+b^{6})\,$