The first part of the question is straight forward and easy to do.  For the second part of the question, you are required to use the first three terms of the binomial expansion from the first part, find a suitable substitute for x  before finding the first three terms of the second expression. ${{\left( {1-3x} \right)}^{5}}=\left( {\begin{array}{*{20}{c}} 5 \\ 0 \end{array}} \right){{\left( 1 \right)}^{5}}{{\left( {-3x} \right)}^{0}}+\left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right){{\left( 1 \right)}^{4}}{{\left( {-3x} \right)}^{1}}+\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){{\left( 1 \right)}^{3}}{{\left( {-3x} \right)}^{2}}+...$ ${{\left( {1-3x} \right)}^{5}}=1-15x+90{{x}^{2}}+...$

Since $1-3x=1-3p-6{{p}^{2}}$ $-3x=-3p-6{{p}^{2}}$ $-3x=-3(p+2{{p}^{2}})$ $x=p+2{{p}^{2}}$

Substitute $x=p+2{{p}^{2}}$ into the equation below ${{\left[ {1-3\left( {p+2{{p}^{2}}} \right)} \right]}^{5}}=1-15(p+2{{p}^{2}})+90{{(p+2{{p}^{2}})}^{2}}+...$ ${{\left[ {1-3p-6{{p}^{2}}} \right]}^{5}}=1-15p-30{{p}^{2}}+90({{p}^{2}}+4{{p}^{3}}+4{{p}^{4}})...$ ${{\left[ {1-3p-6{{p}^{2}}} \right]}^{5}}=1-15p+60{{p}^{2}}...$

Additional Math (AMath)  Group Tuition on Sunday at Woodlands.  Individual Math Tuition available at Choa Chu Kang, Yew Tee, Yishun and Sembawang.