Question 1141476
{{{(x-3y)^5 }}}

The expansion is given by the following formula:

 {{{(a+b)^n}}}={{{(nCk)}}}{{{sum((a^(n-k)*b^k),k=0,n)}}} 




Now, calculate the product for every value of {{{k}}} from {{{0 }}}to {{{5}}}.

{{{k=0}}}: {{{(5C0)(-3y)^5-0(x)^0=5!(5-0)!0!(-3y)^5(x)^0=-243y^5}}}
{{{k=1}}}: {{{(5C1)(-3y)^5-1(x)^1=5!(5-1)!1!(-3y)^4(x)^1=405xy^4}}}
{{{k=2}}}: {{{(5C2)(-3y)^5-2(x)^2=5!(5-2)!2!(-3y)^3(x)^2=-270x^2y^3}}}
{{{k=3}}}: {{{(5C3)(-3y)^5-3(x)^3=5!(5-3)!3!(-3y)^2(x)^3=90x^3y^2}}}
{{{k=4}}}: {{{(5C4)(-3y)^5-4(x)^4=5!(5-4)!4!(-3y)^1(x)^4=-15x^4y}}}
{{{k=5}}}: {{{(5C5)(-3y)^5-5(x)^5=5!(5-5)!5!(-3y)^0(x)^5=x^5}}}


answer: {{{(x-3y)^5 =x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5}}}