Question

$-1\frac{1}{3} (a+b)^{2}$

Answer 1

$-1\frac{1}{3}(a+b)^2=-1\frac{1}{3}a^2-1\frac{1}{3}*2ab-1\frac{1}{3}b^2=-1\frac{1}{3}a^2-\frac{4}{3}*2ab-1\frac{1}{3}b^2=-1\frac{1}{3}a^2-\frac{8}{3}ab-1\frac{1}{3}b^2=-1\frac{1}{3}a^2-2\frac{2}{3}ab-1\frac{1}{3}b^2$