Simplify an expression (HARD)

Algebra Level pending

x 4 ( x + y ) 2 − x 4 y 2 ( x + y ) 4 − y 4 + 4 y 4 ( x + y ) 2 ( x + y ) 4 + y 2 ( x + y ) 2 \frac{{{x^4}{{\left( {x + y} \right)}^2} - {x^4}{y^2}}}{{{{\left( {x + y} \right)}^4} - {y^4}}} + \frac{{4{y^4}{{\left( {x + y} \right)}^2}}}{{{{\left( {x + y} \right)}^4} + {y^2}{{\left( {x + y} \right)}^2}}}

Simplify the expression above, where x x and y y are real numbers and x ≠ { 0 − y − 2 y x \ne \begin{cases} 0 \\ -y \\ -2y \end{cases} .

( x + y ) 4 − ( x − y ) 4 ( x − y ) 4 \frac{{{{\left( {x + y} \right)}^4} - {{\left( {x - y} \right)}^4}}}{{{{\left( {x - y} \right)}^4}}} ( x + y ) 2 + y 2 {\left( {x + y} \right)^2} + {y^2} ( x − y ) 2 + y 2 {\left( {x - y} \right)^2} + {y^2} ( x + y ) 4 {\left( {x + y} \right)^4} ( x + y ) 2 + y 2 ( x − y ) 2 + y 2 \frac{{{{\left( {x + y} \right)}^2} + {y^2}}}{{{{\left( {x - y} \right)}^2} + {y^2}}} 1 ( ( x + y ) 2 − y 2 ) ( ( x + y ) 2 + y 2 ) \left( {{{\left( {x + y} \right)}^2} - {y^2}} \right)\left( {{{\left( {x + y} \right)}^2} + {y^2}} \right)

Tomáš Hauser by Tomáš Hauser , 21, Czech Republic

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2 solutions

Chew-Seong Cheong
Jan 29, 2020

X = x 4 ( x + y ) 2 − x 4 y 2 ( x + y ) 4 − y 4 + 4 y 4 ( x + y ) 2 ( x + y ) 4 + y 2 ( x + y ) 2 = x 4 ( ( x + y ) 2 − y 2 ) ( ( x + y ) 2 + y 2 ) ( ( x + y ) 2 − y 2 ) + 4 y 4 ( x + y ) 2 ( x + y ) 2 ( ( x + y ) 2 + y 2 ) = x 4 + 4 y 4 ( x + y ) 2 + y 2 By Sophie Germain identity = ( ( x + y ) 2 + y 2 ) ( ( x − y ) 2 + y 2 ) ( x + y ) 2 + y 2 = ( x − y ) 2 + y 2 \begin{aligned} X & = \frac {x^4(x+y)^2-x^4y^2}{(x+y)^4-y^4} + \frac {4y^4(x+y)^2}{(x+y)^4+y^2(x+y)^2} \\ & = \frac {x^4 \cancel{\left((x+y)^2-y^2\right)}}{\left((x+y)^2+y^2\right) \cancel{\left((x+y)^2-y^2\right)}} + \frac {4y^4 \cancel{(x+y)^2}}{\cancel{(x+y)^2} \left((x+y)^2+y^2\right)} \\ & = \frac {\blue{x^4+4y^4}}{(x+y)^2+y^2} & \small \blue{\text{By Sophie Germain identity}} \\ & = \frac {\blue{\left((x+y)^2+y^2\right)\cancel{\left((x-y)^2+y^2\right)}}}{\cancel{(x+y)^2+y^2}} \\ & = \boxed{(x-y)^2+y^2} \end{aligned}

Reference: Sophie Germain identity

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Hey Chew-Seong, how do you get those crossed-terms to appear in LATEX? Thanks!

tom engelsman - 1 year, 4 months ago

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First you have to declare \require{cancel}. Then you can use it as \cancel{E=mc^2} E = m c 2 \cancel{E=mc^2} .

Chew-Seong Cheong - 1 year, 4 months ago
Tom Engelsman
Jan 29, 2020

The above expression can be factorized according to:

x 4 ( ( x + y ) 2 − y 2 ) ( ( x + y ) 2 + y 2 ) ( ( x + y ) 2 − y 2 ) + 4 y 2 ( x + y ) 2 ( x + y ) 2 ( ( x + y ) 2 + y 2 ) \frac{x^4((x+y)^2 - y^2)}{((x+y)^2 + y^2)((x+y)^2 - y^2)} + \frac{4y^2(x+y)^2}{(x+y)^2((x+y)^2 + y^2)} ;

or x 4 ( x + y ) 2 + y 2 + 4 y 4 ( x + y ) 2 + y 2 ; \frac{x^4}{(x+y)^2 + y^2} + \frac{4y^4}{(x+y)^2 + y^2};

or x 4 + 4 y 4 ( x + y ) 2 + y 2 ; \frac{x^4 + 4y^4}{(x+y)^2 + y^2};

or ( x 2 + 2 x y + 2 y 2 ) ( x 2 − 2 x y + 2 y 2 ) ( x + y ) 2 + y 2 ; \frac{(x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)}{(x+y)^2 + y^2};

or ( ( x + y ) 2 + y 2 ) ( ( x − y ) 2 + y 2 ) ( x + y ) 2 + y 2 ; \frac{((x+y)^2 + y^2)((x-y)^2 + y^2)}{(x+y)^2 + y^2};

or ( x − y ) 2 + y 2 . \boxed{(x-y)^2 + y^2}.

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