An algebra problem by Wildan Bagus Wicaksono

Relevant wiki: Rules of Exponents

$=\frac { { \left( { 3 }^{ 2017 } \right) }^{ 2 }-{ \left( { 3 }^{ 2015 } \right) }^{ 2 } }{ { \left( { 3 }^{ 2018 } \right) }^{ 2 }-{ \left( { 3 }^{ 2016 } \right) }^{ 2 } } \\ =\frac { { 3 }^{ 4034 }-{ 3 }^{ 4030 } }{ { 3 }^{ 4036 }-{ 3 }^{ 4032 } } \\ =\frac { { 3 }^{ 4034 }-{ 3 }^{ 4030 } }{ { 3 }^{ 2 }\left( { 3 }^{ 4034 }-{ 3 }^{ 4030 } \right) } \\ =\frac { 1 }{ 9 }$

$Let \hskip 0.5em u = 2015 \hskip 0.8em$ And then plug into the expression: $= \frac{ { \left ( {3} ^ { u+ 2 } \right) } ^ {2} - { \left( {3}^{u} \right) }^ {2} } {{ \left ( {3} ^ { u+ 3 } \right) } ^ {2} - { \left ( {3} ^ { u + 1 } \right) } ^ {2} } = \frac{ {3} ^ {2u} \times {3}^ {4} - {3} ^{2u} } { {3} ^ {2u} \times {3}^ {6} - {3} ^ {2u} \times {3}^ {2} } \\[0.5in] = \, \frac { {3} ^ {2u} \left ( {3} ^ {4} - 1 \right) } { {3} ^ {2u} \left ( 3 ^ 6 - 3 ^ 2 \right) } = \frac {3 ^ 4 - 1} {3 ^ {2 + 4} - 3 ^ 2} \\[0.5in] = \frac { 3^4 -1 } {3^2 \times 3^4 - 3^2} = \frac { \left(3^4 - 1 \right)} {3^2 \left( 3^4 - 1 \right) } \\[0.5in] = \frac {1} {3^2} = \, \boxed{ \dfrac {1} {9} }$

$\begin{aligned} \frac {\left(3^{2017}\right)^2-\left(3^{2015}\right)^2}{\left(3^{2018}\right)^2-\left(3^{2016}\right)^2} & = \frac {3^{4034}- 3^{4030}}{3^{4036}- 3^{4032}} = \frac {3^{4030}\left(3^4- 1\right)}{3^{4032}\left(3^4- 1\right)} = \frac 1{3^2} = \boxed{\dfrac 19}\end{aligned}$

$\frac{(3^{2017})^2 - (3^{2015})^2}{(3^{2018})^2 - (3^{2016})^2}$ $= \frac{(3^{2017})^2 - (3^{2015})^2}{3^2*(3^{2017})^2 - 3^2*(3^{2015})^2}$ $= \frac{1}{3^2} * \frac{(3^{2017})^2 - (3^{2015})^2}{(3^{2017})^2 - (3^{2015})^2}$ $= \frac{1}{3^2} = \frac{1}{9}$

I'm a firm believer in avoiding unnecessary computations.

Initial expression:

$\frac{(3^{2017})^2-(3^{2015})^2}{(3^{2018})^2-(3^{2016})^2}$

Apply Difference of Squares

$\frac{(3^{2017}+3^{2015})(3^{2017}-3^{2015})}{(3^{2018}+3^{2016})(3^{2018}-3^{2016})}$

Divide each term by $3^{2015}$

$\frac{3^{2015}(3^{2}+3^{0})(3^{2}-3^{0})}{3^{2015}(3^{3}+3^{1})(3^{3}-3^{1})}$

Simplify

$\frac{(9+1)(9-1)}{(27+3)(27-3)}$

Simplify, Simplify

$\frac{(10)(8)}{(30)(24)}$ = $\frac{(1)(1)}{(3)(3)}$ = $\frac{1}{9}$

Take (3^2015)^2 common on numerator and denominator then over the both you'll get (81➖ 1)➗ (729➖ 9)= 80/720= 1/9

LaTex: $\huge \text{We notice that the common factor is : } \; \left (3^{2015}\right )^{2}$

LaTex: $\Huge \frac{\left(3^{2017}\right)^2 - \left(3^{2015}\right)^2} {\left (3^{2018}\right )^2 - \left(3^{2016}\right)^2} = \frac{\left (3^{2015}\right)^2 \cdot\ \left(3^4\right) - \left(3^{2015}\right)^2} {\left (3^{2015}\right )^2 \cdot\ \left(3^6\right) - \left(3^{2015}\right)^2 \cdot\ \left(3^2\right)}$

LaTex: $\Huge \frac{\left(3^{2017}\right)^2 - \left(3^{2015}\right)^2} {\left (3^{2018}\right )^2 - \left(3^{2016}\right)^2} = \frac{\left(3^{2015}\right )^2 \cdot\ \left(\left(3^4\right) - 1\right)} { \left(3^{2015}\right )^2 \cdot\ \left(\left(3^4\right) - 1 \right) \cdot\ \left(3^2\right)}$

LaTex: $\Huge \frac{\left(3^{2017}\right)^2 - \left(3^{2015}\right)^2} {\left (3^{2018}\right )^2 - \left(3^{2016}\right)^2} = \frac{1}{3^2} = \color{#D61F06}{\boxed{ \frac{1}{9}}}$

[(3^2017)²-(3^2015)²]/[(3^2018)²-(3^2016)²] a²-b²= (a+b)(a-b) =[(3^2017+3^2015)(3^2017-3^2015)] / [(3^2018+3^2016)(3^2018-3^2016)] =[3^2015(3²+1)3^2015(3²-1)] / [3^2016(3²+1)3^2016(3^2-1) Notebook = 3²-1 and 3²+1 habis dibagi =3^2015+2015-2016-2016 =3^-2 =1/3² =1/9 Just it.. Thank you 😀

If you take a factor of $3^2$ out of the denominator, you end up with the numerator.

Put 3^2015=k by substituting in the expression( (k×9)^2-k^2)÷((k×27)^2-(k×3)^2) expression=(81-1)÷(27^2-3^2)=80÷(30×24)=1÷9##