y^2/3 + 3y^-1 -2 when y =1/8
You asked:
Calculate the value of \(\frac{{y}^{2}}{3} + 3 {y}^{-1} - 2\) when \(y = \frac{1}{8}\).
MathBot Answer:
\[\frac{{y}^{2}}{3} + 3 {y}^{-1} - 2 = \frac{4225}{192}\]
$$\begin{aligned}\frac{y^{2}}{3} - 2 + \frac{3}{y}&=\frac{\left(\frac{1}{8}\right)^{2}}{3} + \frac{3}{\frac{1}{8}} - 2\\&=3 \cdot 8 + \frac{1}{3 \cdot 8^{2}} - 2\\&=3 \cdot 8 + \frac{\frac{1}{3}}{64} - 2\\&=3 \cdot 8 + \frac{1}{192} - 2\\&=\frac{4225}{192}\end{aligned}$$