To expand a binomial of the form #(x+a)^n#, we use the Binomial Theorem.
The binomial theorem is as follows:
#sum_(k=0)^n((n),(k))x^(n-k)a^k#, where #((n),(k))=(n!)/(k!(n-k)!)#
Here Pascal's Triangle comes into play. What I wrote above as #((n),(k))#, (read as "n choose k") corresponds to the row and column number of a coefficient.
#n# is the row number, and #k# the column number.
Use the binomial theorem where #x=3x,a=-1,n=5#:
#sum_(k=0)^5((5),(k))(3x)^(5-k)(-1)^k#.
When #k=0#,
#(5!)/(0!(5-0)!)*(3x)^(5-0)(-1)^0#
#=1*243x^5*1#
#=243x^5#
When #k=1#,
#(5!)/(1!(5-1)!)*(3x)^(5-1)(-1)^1#
#=5*81x^4*-1#
#=-405x^4#
When #k=2#,
#(5!)/(2!(5-2)!)*(3x)^(5-2)(-1)^2#
#=10*27x^3*1#
#=270x^3#
When #k=3#,
#(5!)/(3!(5-3)!)*(3x)^(5-3)(-1)^3#
#=10*9x^2*-1#
#=-90x^2#
When #k=4#,
#(5!)/(4!(5-4)!)*(3x)^(5-4)(-1)^4#
#=5*3x*1#
#=15x#
When #k=5#
#(5!)/(5!(5-5)!)*(3x)^(5-5)(-1)^5#
#=1*1*-1#
#=-1#
Add up all your answers, and you get:
#243x^5-405x^4+270x^3-90x^2+15x-1#
Which is your answer.
