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

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

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Answer 1 · 2017-12-05T05:30:02

$\frac{x^{2} - 5 x + 5}{x^{3} + 9 x^{2} + 27 x - 27} = \frac{x^{2} - 5 x + 5}{{(x - 3)}^{3}}$ $(x^2 -5x +5)/(x^3 +9x^2 + 27x- 27) =(x^2 -5x +5)/(x-3)^3$

Explanation:

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

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

${(x - 3)}^{2} = x^{2} - 6 x + 9$ $(x-3)^2 = x^2 -6x +9$ ------------ = $a^{2}$ $a^2$ and
${(x - 3)}^{3} = x^{3} - 27 - 3 \times x^{2} (3) + 3 \times \times x (9)$ $(x-3)^3 = x^3 - 27 - 3xx x^2 (3) + 3xxxxx(9)$

${(x - 3)}^{3} = x^{3} - 27 - 9 \times x^{2} + 27 x$ $(x-3)^3 = x^3 - 27 -9xx x^2 + 27x$

${(x - 3)}^{3} = x^{3} + 9 x^{2} + 27 x - 27$ $(x-3)^3 = x^3 +9x^2 + 27x- 27$ ------ $a^{3}$ $a^3$

And, given expression can be written as:

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

Now solve by equating the denominators:

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

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

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

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

Substitute values:

$\Rightarrow \frac{x^{2} - 6 x + 9 + x - 3 - 1}{x^{3} + 9 x^{2} + 27 x - 27}$ $=>(x^2 -6x +9 + x-3 -1)/(x^3 +9x^2 + 27x- 27)$

$\Rightarrow \frac{x^{2} - 5 x + 5}{x^{3} + 9 x^{2} + 27 x - 27}$ $=> (x^2 -5x +5)/(x^3 +9x^2 + 27x- 27)$

$\Rightarrow \frac{x^{2} - 5 x + 5}{{(x - 3)}^{3}}$ $=> (x^2 -5x +5)/ (x-3)^3$

Answer 2 · 2017-12-05T09:13:38

$\frac{x^{2} - 5 x + 5}{{(x - 3)}^{3}}$ $(x^2-5x+5)/(x-3)^3$

Explanation:

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

$the common denominator of$ $"the common denominator of "$

$(x - 3), {(x - 3)}^{2} and {(x - 3)}^{3} is {(x - 3)}^{3}$ $(x-3),(x-3)^2" and "(x-3)^3" is "(x-3)^3$

$multiply numerator/denominator of$ $"multiply numerator/denominator of "$

$\frac{1}{x - 3} by {(x - 3)}^{2}$ $1/(x-3)" by "(x-3)^2$

$\Rightarrow \frac{1}{x - 3} \times \frac{{(x - 3)}^{2}}{{(x - 3)}^{2}} = \frac{{(x - 3)}^{2}}{{(x - 3)}^{3}}$ $rArr1/(x-3)xx(x-3)^2/(x-3)^2=(x-3)^2/(x-3)^3$

$multiply numerator/denominator of$ $"multiply numerator/denominator of"$

$\frac{1}{{(x - 3)}^{2}} by (x - 3)$ $1/(x-3)^2" by "(x-3)$

$\Rightarrow \frac{1}{{(x - 3)}^{2}} \times \frac{x - 3}{x - 3} = \frac{x - 3}{{(x - 3)}^{3}}$ $rArr1/(x-3)^2xx(x-3)/(x-3)=(x-3)/(x-3)^3$

$putting this together gives$ $"putting this together gives"$

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

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

$= \frac{x^{2} - 6 x + 9 + x - 3 - 1}{{(x - 3)}^{3}}$ $=(x^2-6x+9+x-3-1)/(x-3)^3$

$= \frac{x^{2} - 5 x + 5}{{(x - 3)}^{3}} \to (x \neq 3)$ $=(x^2-5x+5)/(x-3)^3to(x!=3)$

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