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

2 Answers
Dec 5, 2017

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

Explanation:

\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

Dec 5, 2017

(x^2-5x+5)/(x-3)^3

Explanation:

"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)