How do you combine #\frac { 1} { x - 3} + \frac { 1} { ( x - 3) ^ { 2} } - \frac { 1} { ( x - 3) ^ { 3} }# into one term?

#\frac { 1} { x - 3} + \frac { 1} { ( x - 3) ^ { 2} } - \frac { 1} { ( x - 3) ^ { 3} }#

Take #(x-3)= a#, So,

#(x-3)^2 = x^2 -6x +9# ------------ =#a^2# and
#(x-3)^3 = x^3 - 27 - 3xx x^2 (3) + 3xxxxx(9)#

#(x-3)^3 = x^3 - 27 -9xx x^2 + 27x#

#(x-3)^3 = x^3 +9x^2 + 27x- 27#------#a^3#

And, given expression can be written as:

#=>\frac { 1} { a} + \frac { 1} { a^ { 2} } - \frac { 1} { a ^ { 3} }#

Now solve by equating the denominators:

#=>\frac { 1} { a} + \frac { 1} { a^ { 2} } - \frac { 1} { a ^ { 3} }#

#=> \frac { 1} { a} (a^2/a^2) + \frac { 1} { a^ { 2} }(a/a) - \frac { 1} { a ^ { 3} }#

#=> \frac { a^2} { a^3} + \frac { a} { a^ { 3} } - \frac { 1} { a ^ { 3} }#

#=> \frac { a^2 +a -1} { a^3} #

Substitute values:

#=>(x^2 -6x +9 + x-3 -1)/(x^3 +9x^2 + 27x- 27)#

#=> (x^2 -5x +5)/(x^3 +9x^2 + 27x- 27)#

#=> (x^2 -5x +5)/ (x-3)^3#

#"we require the fractions to have a "color(blue)"common denominator"#

#"the common denominator of "#

#(x-3),(x-3)^2" and "(x-3)^3" is "(x-3)^3#

#"multiply numerator/denominator of "#

#1/(x-3)" by "(x-3)^2#

#rArr1/(x-3)xx(x-3)^2/(x-3)^2=(x-3)^2/(x-3)^3#

#"multiply numerator/denominator of"#

#1/(x-3)^2" by "(x-3)#

#rArr1/(x-3)^2xx(x-3)/(x-3)=(x-3)/(x-3)^3#

#"putting this together gives"#

#(x-3)^2/(x-3)^3+(x-3)/(x-3)^3-1/(x-3)^3#

#"the fractions have a common denominator so add"#
#"the numerators leaving the common denominator"#

#=(x^2-6x+9+x-3-1)/(x-3)^3#

#=(x^2-5x+5)/(x-3)^3to(x!=3)#