Exercise:
Let $h\left( x \right)=1+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{6}}}{6!}+....\,\,\,\,\,,\,\,\,\,\,q\left( x \right)=x+\frac{{{x}^{4}}}{4!}+\frac{{{x}^{7}}}{7!}+...\,\,\,\,,\,\,\,\,p\left( x \right)=\frac{{{x}^{2}}}{2!}+\frac{{{x}^{5}}}{5!}+\frac{{{x}^{8}}}{8!}+...$
Show that , ${{h}^{3}}\left( x \right)+{{q}^{3}}\left( x \right)+{{p}^{3}}\left( x \right)-3h\left( x \right)q\left( x \right)p\left( x \right)=1$
Solution: observe that $h\left( x \right)+q\left( x \right)+p\left( x \right)=1+x+\frac{{{x}^{2}}}{2!}+...={{e}^{x}}$
We have ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( {{a}^{3}}+{{b}^{3}} \right)+{{c}^{3}}$
And we know that ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$
So ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)+{{c}^{3}}={{\left( a+b \right)}^{3}}+{{c}^{3}}-3ab\left( a+b \right)$
But ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)$
So ${{\left( a+b \right)}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{\left( a+b \right)}^{2}}+{{c}^{2}}-c\left( a+b \right) \right)$
Hence ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-bc-ac \right)-3ab\left( a+b \right)$
So ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-ac-bc \right)-3ab\left( a+b \right)-3abc$
$=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-bc-ac \right)-3ab\left( a+b+c \right)$
$=\left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-bc-ac-3ab \right]$
$=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)$
Thus ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-ac-bc \right)$
So ${{h}^{3}}\left( x \right)+{{q}^{3}}\left( x \right)+{{p}^{3}}\left( x \right)-3h\left( x \right)q\left( x \right)p\left( x \right)$
$=\left( h\left( x \right)+q\left( x \right)+p\left( x \right) \right)\left( {{h}^{2}}\left( x \right)+{{q}^{2}}\left( x \right)+{{p}^{2}}\left( x \right)-h\left( x \right)q\left( x \right)-h\left( x \right)p\left( x \right)-q\left( x \right)p\left( x \right) \right)$
$={{e}^{x}}\left( {{h}^{2}}\left( x \right)+{{q}^{2}}\left( x \right)+{{p}^{2}}\left( x \right)-h\left( x \right)q\left( x \right)-h\left( x \right)p\left( x \right)-q\left( x \right)p\left( x \right) \right)$
Let $z\left( x \right)={{h}^{2}}\left( x \right)+{{q}^{2}}\left( x \right)+{{p}^{2}}\left( x \right)-h\left( x \right)q\left( x \right)-h\left( x \right)p\left( x \right)-q\left( x \right)p\left( x \right)$
So $z'\left( x \right)=2\left( hh'+qq'+pp' \right)-\left( h'q+q'h \right)-\left( h'p+p'h \right)-\left( q'p+p'q \right)$
But $\frac{d}{dx}\left( h\left( x \right) \right)=0+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{5}}}{5!}+....=p\left( x \right)$ , $\frac{d}{dx}\left( q\left( x \right) \right)=1+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{6}}}{6!}+....=h\left( x \right)$
And $\frac{d}{dx}\left( p\left( x \right) \right)=x+\frac{{{x}^{4}}}{4!}+....=q\left( x \right)$
So $z'\left( x \right)=2\left( hp+qh+pq \right)-\left( pq+{{h}^{2}} \right)-\left( {{p}^{2}}+qh \right)-\left( hp+{{q}^{2}} \right)$
$=\left( hp+qh+pq \right)-\left( {{h}^{2}}+{{p}^{2}}+{{q}^{2}} \right)=-z\left( x \right)$
Since \[z'\left( x \right)=-z\left( x \right)\Leftrightarrow \frac{dz}{dx}=-z\Leftrightarrow \int{\frac{dz}{z}}=-\int{dx}\Leftrightarrow \ln \left| z \right|=-x+c\Leftrightarrow z\left( x \right)=c{{e}^{-x}}\]
Observe that $h\left( 0 \right)=1\,\,,\,\,q\left( 0 \right)=0\,\,\And \,\,p\left( 0 \right)=0\Leftrightarrow z\left( 0 \right)=1$ hence $c=1$
Therefore ${{h}^{3}}\left( x \right)+{{q}^{3}}\left( x \right)+{{p}^{3}}\left( x \right)-3h\left( x \right)q\left( x \right)p\left( x \right)={{e}^{x}}{{e}^{-x}}={{e}^{0}}=1$ Q.E.D
Or: by differentiating both sides by $x$ to get
So $3{{h}^{2}}h'+3{{q}^{2}}q'+3{{p}^{2}}p'-3\left( h'\left( qp \right)+\left( qp \right)'h \right)=0$
$\Leftrightarrow 3{{h}^{2}}q+3{{q}^{2}}h+3{{p}^{2}}q-3\left( {{p}^{2}}q+\left( q'p+p'q \right)h \right)=3{{h}^{2}}q+3{{q}^{2}}h+3{{p}^{2}}q-3\left( {{p}^{2}}q+{{h}^{2}}p+{{q}^{2}}h \right)=0$
so $f'\left( x \right)=0\Leftrightarrow f\left( x \right)=c$
Now to find the value of $c$ plug $x=0$ to get $c=1$ Q.E.D
*____________________________
first Method credit to شحات جامع
2nd Method credit to Kunihiko Chikaya
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