Question

If $\text{P}\left(\text{y}\right)\text{=}{\text{y}}^{\text{2}}\text{- 3}\sqrt{\text{2y}}\text{+ 1}$ then find $\text{P}\left(\sqrt[\text{3}]{\text{3}\text{2}}\right)$

Answer 1

Given

$P\left(y\right)=y^2-3\sqrt{2y}+1$

$P\left(\sqrt[3]{32}\right)=(3\sqrt{3}{)}^2-3\sqrt{2.(2{)}^{\frac{1}{2}}}+1$

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

$=2^{\frac{2}{3}}-3^{\frac{4}{3}\times \frac{1}{2}}+1$

$=2^{\frac{2}{3}}-{3.2}^{\frac{2}{3}}+1$

$=-{2.2}^{\frac{2}{3}}+1$

$=-2^{\frac{2}{3}+1}+1$

$=1-2^{\frac{5}{3}}$