text: In this case make the substitution $y=x+4$, which transforms the equation into
  $$(y-1)^4+(y+1)^4=20.$$
  This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'.
  
  If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur:
  $$(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.$$
  So, the odd factors cancel to give
  $$2y^4+12y^2-18=0$$
  This then becomes the simple equation
  $$y^4+6y^2-9=0\,,$$
  which is a quadratic equation in $y^2$ with solutions
  $$y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2).$$
  One of these solutions is positive; taking the square root gives two real values for $y$ as
  $$y = \pm \sqrt{3\left(\sqrt{2}-1)\right)}$$
  Therefore two real solutions to the equation are
  $$x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)}$$
explanation: The document contains a detailed mathematical derivation and solution to a specific equation. It starts with a substitution, simplifies the equation, and ultimately finds the real solutions for \( x \). The text does not contain a question but rather presents a step-by-step solution to a problem.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: In this case make the substitution $y=x+4$, which transforms the equation into
  $$(y-1)^4+(y+1)^4=20.$$
  This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'.
  
  If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur:
  $$(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.$$
  So, the odd factors cancel to give
  $$2y^4+12y^2-18=0$$
  This then becomes the simple equation
  $$y^4+6y^2-9=0\,,$$
  which is a quadratic equation in $y^2$ with solutions
  $$y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2).$$
  One of these solutions is positive; taking the square root gives two real values for $y$ as
  $$y = \pm \sqrt{3\left(\sqrt{2}-1)\right)}$$
  Therefore two real solutions to the equation are
  $$x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)}$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed mathematical derivation and solution to a specific equation. It starts with a substitution, simplifies the equation, and ultimately finds the real solutions for \( x \). The text does not contain a question but rather presents a step-by-step solution to a problem.
  
  Label: no-question