x^3+y^3+x^2+4y^2+6
put x = (-2/3) and y=(-8/3)
You asked:
Calculate the value of \({x}^{3} + {y}^{3} + {x}^{2} + 4 \cdot {y}^{2} + 6\) when \(x = \frac{-2}{3}\) and \(y = \frac{-8}{3}\).
MathBot Answer:
\[{x}^{3} + {y}^{3} + {x}^{2} + 4 \cdot {y}^{2} + 6 = {\left( \frac{-2}{3} \right)}^{3} + {\left( \frac{-8}{3} \right)}^{3} + {\left( \frac{-2}{3} \right)}^{2} + 4 \cdot {\left( \frac{-8}{3} \right)}^{2} + 6 = \frac{422}{27}\]