(x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?(x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?\frac{{{{\left( {x - y} \right)}^3} + {{\left( {y - z} \right)}^3} + {{\left( {z - x} \right)}^3}}}{{{{\left( {{x^2} - {y^2}} \right)}^3} + {{\left( {{y^2} - {z^2}} \right)}^3} + {{\left( {{z^2} - {x^2}} \right)}^3}}}{\text{ is = ?}}" /> (x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?(x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?\frac{{{{\left( {x - y} \right)}^3} + {{\left( {y - z} \right)}^3} + {{\left( {z - x} \right)}^3}}}{{{{\left( {{x^2} - {y^2}} \right)}^3} + {{\left( {{y^2} - {z^2}} \right)}^3} + {{\left( {{z^2} - {x^2}} \right)}^3}}}{\text{ is = ?}}" /> (x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?(x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?\frac{{{{\left( {x - y} \right)}^3} + {{\left( {y - z} \right)}^3} + {{\left( {z - x} \right)}^3}}}{{{{\left( {{x^2} - {y^2}} \right)}^3} + {{\left( {{y^2} - {z^2}} \right)}^3} + {{\left( {{z^2} - {x^2}} \right)}^3}}}{\text{ is = ?}}" />

The value of
(x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?
  • 10
  • 21
  • 3
    [2(x+y+z)]−1
  • 4
    [(x+y)(y+z)(z+x)]−1
Answer:- 1
Explanation:-

Solution:
Since (x−y)+(y−z)+(z−x)=0So,(x−y)3+(y−z)3+(z−x)3=3(x−y)(y−z)(z−x)Since(x2−y2)+(y2−z2)(z2−x2)=0So,(x2−y2)+(y2−z2)(z2−x2)=3(x2−y2)(y2−z2)(z2−x2)∴Given expression  = 3(x−y)(y−z)(z−x)3(x2−y2)(y2−z2)(z2−x2)=1(x+y)(y+z)(z+x)=[(x+y)(y+z)(z+x)]−1

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", "text": "The value of
(x−y)3+(y−z)3+(z−x)3(x2−y2)3+(y2−z2)3+(z2−x2)3 is = ?
", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "
Solution:
Since (x−y)+(y−z)+(z−x)=0So,(x−y)3+(y−z)3+(z−x)3=3(x−y)(y−z)(z−x)Since(x2−y2)+(y2−z2)(z2−x2)=0So,(x2−y2)+(y2−z2)(z2−x2)=3(x2−y2)(y2−z2)(z2−x2)∴Given expression  = 3(x−y)(y−z)(z−x)3(x2−y2)(y2−z2)(z2−x2)=1(x+y)(y+z)(z+x)=[(x+y)(y+z)(z+x)]−1
", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "
Solution:
Since (x−y)+(y−z)+(z−x)=0So,(x−y)3+(y−z)3+(z−x)3=3(x−y)(y−z)(z−x)Since(x2−y2)+(y2−z2)(z2−x2)=0So,(x2−y2)+(y2−z2)(z2−x2)=3(x2−y2)(y2−z2)(z2−x2)∴Given expression  = 3(x−y)(y−z)(z−x)3(x2−y2)(y2−z2)(z2−x2)=1(x+y)(y+z)(z+x)=[(x+y)(y+z)(z+x)]−1
", "dateCreated": "7/24/2019 10:09:12 AM" } }
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