Answer with Step-by-step explanation:
We are given that
[tex]3\times (2\div 3)^2+(2\div 3)^2-20\times 2\div 3+12[/tex]
[tex]3\times (\frac{2}{3})^3+(\frac{2}{3})^2-20\times \frac{2}{3}+12[/tex]
[tex]3\times \frac{8}{27}+\frac{4}{9}-\frac{40}{3}+12[/tex]
[tex]\frac{24}{27}+\frac{4}{9}-\frac{40}{3}+12[/tex]
[tex]\frac{24+12}{27}-\frac{40}{3}+12[/tex]
[tex]\frac{36}{27}-\frac{40}{3}+12[/tex]
[tex]\frac{4}{3}-\frac{40}{3}+12[/tex]
[tex]\frac{4-40}{3}+12[/tex]
[tex]-\frac{36}{3}+12[/tex]
[tex]-12+12[/tex]
[tex]=0[/tex]
