Answer
Using the formula for the cross product, we obtain
$\left( {\lambda {\bf{v}}} \right) \times {\bf{w}} = \lambda \left( {{\bf{v}} \times {\bf{w}}} \right)$
Work Step by Step
Let the components of ${\bf{v}}$ and ${\bf{w}}$ be given by ${\bf{v}} = \left( {{v_1},{v_2},{v_3}} \right)$ and ${\bf{w}} = \left( {{w_1},{w_2},{w_3}} \right)$, respectively. So, $\lambda {\bf{v}} = \left( {\lambda {v_1},\lambda {v_2},\lambda {v_3}} \right)$.
Using the formula for the cross product, we have
$\left( {\lambda {\bf{v}}} \right) \times {\bf{w}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\lambda {v_1}}&{\lambda {v_2}}&{\lambda {v_3}}\\
{{w_1}}&{{w_2}}&{{w_3}}
\end{array}} \right|$
$\left( {\lambda {\bf{v}}} \right) \times {\bf{w}} = \left| {\begin{array}{*{20}{c}}
{\lambda {v_2}}&{\lambda {v_3}}\\
{{w_2}}&{{w_3}}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
{\lambda {v_1}}&{\lambda {v_3}}\\
{{w_1}}&{{w_3}}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
{\lambda {v_1}}&{\lambda {v_2}}\\
{{w_1}}&{{w_2}}
\end{array}} \right|{\bf{k}}$
$\left( {\lambda {\bf{v}}} \right) \times {\bf{w}} = \left( {\lambda {v_2}{w_3} - \lambda {v_3}{w_2}} \right){\bf{i}} - \left( {\lambda {v_1}{w_3} - \lambda {v_3}{w_1}} \right){\bf{j}} + \left( {\lambda {v_1}{w_2} - \lambda {v_2}{w_1}} \right){\bf{k}}$
$\left( {\lambda {\bf{v}}} \right) \times {\bf{w}} = \lambda \left( {\left( {{v_2}{w_3} - {v_3}{w_2}} \right){\bf{i}} - \left( {{v_1}{w_3} - {v_3}{w_1}} \right){\bf{j}} + \left( {{v_1}{w_2} - {v_2}{w_1}} \right){\bf{k}}} \right)$
From Exercise 49, we have
${\bf{v}} \times {\bf{w}} = \left( {{v_2}{w_3} - {v_3}{w_2}} \right){\bf{i}} - \left( {{v_1}{w_3} - {v_3}{w_1}} \right){\bf{j}} + \left( {{v_1}{w_2} - {v_2}{w_1}} \right){\bf{k}}$
Therefore,
$\left( {\lambda {\bf{v}}} \right) \times {\bf{w}} = \lambda \left( {{\bf{v}} \times {\bf{w}}} \right)$
