Chapter 8, Problem 7P

To determine

To show: That $u^3+v^3+w^3-3uvw=1$ .

Answer to Problem 7P

It is proved that $u^3+v^3+w^3-3uvw=1$ .

Explanation of Solution

Given:

The given equations are as follows.

  $\begin{array}{c}u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots \\ v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\cdots \\ w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots \end{array}$

Calculation:

Differentiate the equation $u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots$ with respect to $x$ .

  $\begin{array}{c}\frac{du}{dx}=\frac{d}{dx}\left(1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots \right)\\ u\text{'}=\frac{3x^2}{3!}+\frac{6x^5}{6!}+\frac{9x^8}{9!}+\cdots \\ =\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots \\ =w\end{array}$

Differentiate the equation $v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\cdots$ with respect to $x$ .

  $\begin{array}{c}\frac{dv}{dx}=\frac{d}{dx}\left(x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\cdots \right)\\ v\text{'}=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots \\ =u\end{array}$

Differentiate the equation $w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots$ with respect to $x$ .

  $\begin{array}{c}\frac{dw}{dx}=\frac{d}{dx}\left(\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots \right)\\ w\text{'}=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\cdots \\ =v\end{array}$

In order to show that $u^3+v^3+w^3-3uvw=C$ , take the left hand side term and differentiate.

Differentiate the left hand side term with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(u^3+v^3+w^3-3uvw\right)=3u^{2}u\text{'}+3v^{2}v\text{'}+3w^{2}w\text{'}-3\left(u\text{'}vw+uv\text{'}w+uvw\text{'}\right)\\ =3u^{2}w+3v^{2}u+3w^{2}v-3\left(v{w}^2+u^{2}w+u{v}^2\right)\\ =0\end{array}$

Integrate the equation on both sides.

  $\begin{array}{c}{\displaystyle \int \frac{d}{dx}\left(u^3+v^3+w^3-3uvw\right)}dx={\displaystyle \int 0dx}\\ u^3+v^3+w^3-3uvw=C\end{array}$

Substitute 0 for $x$ in the equation given equations to find the values of $u$ , $v$ , and $w$ .

  $\begin{array}{c}u=1\\ v=0\\ w=0\end{array}$

Substitute 1 for $u$ , 0 for $v$ , and 0 for $w$ in the equation $u^3+v^3+w^3-3uvw=C$ to find the value of $C$ .

  $\begin{array}{c}{\left(1\right)}^3+{\left(0\right)}^3+{\left(0\right)}^3-3\left(1\right)\left(0\right)\left(0\right)=C\\ C=1\end{array}$

Substitute 1 for $C$ in the equation $u^3+v^3+w^3-3uvw=C$ .

  $u^3+v^3+w^3-3uvw=1$

Therefore, it is proved that $u^3+v^3+w^3-3uvw=1$ .