Question

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

Answer 1

`(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`

Answer 2

`(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)`