Answer
$f\left( x \right) = \cos \left( {{x^3}} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{6n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^6}}}{{2!}} + \dfrac{{{x^{12}}}}{{4!}} - \dfrac{{{x^{18}}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
${f^{\left( 6 \right)}}\left( 0 \right) = - 360$
Work Step by Step
From Table 2 we have
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
Substituting ${x^3}$ for $x$ in the series above we get
$f\left( x \right) = \cos \left( {{x^3}} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{6n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^6}}}{{2!}} + \dfrac{{{x^{12}}}}{{4!}} - \dfrac{{{x^{18}}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
According to Maclaurin series: $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$.
Comparing with the expansion above, we get
$\dfrac{{{f^{\left( 6 \right)}}\left( 0 \right)}}{{6!}} = - \dfrac{1}{{2!}}$
So,
${f^{\left( 6 \right)}}\left( 0 \right) = - \dfrac{{6!}}{{2!}} = - 3\cdot4\cdot5\cdot6 = - 360$
