Question

Find the value of the expressions 3x^3-2y^3-6x^2y^2+xy for x=2/3 and y=1/2

Answer 1

Hi!

$3x^3-2y^3-6x^2y^2+xy=\\\\=3\cdot(\frac{2}{3})^3-2\cdot(\frac{1}{2})^3-6\cdot(\frac{2}{3})^2\cdot(\frac{1}{2})^2+\frac{2}{3}\cdot\frac{1}{2}=\\\\=3\cdot\frac{8}{27}-2\cdot\frac{1}{8}-6\cdot\frac{4}{9}\cdot\frac{1}{4}+\frac{1}{3}=\\\\=\frac{8}{9}-\frac{1}{4}-\frac{2}{3}+\frac{1}{3}=\frac{8}{9}-\frac{1}{4}-\frac{1}{3}=\frac{32}{36}-\frac{9}{36}-\frac{12}{36}=\boxed{\frac{11}{36}}$