Question:

What is the value of \((x^4+y^4+x^2y^2)\frac{x^2-2xy+y^2}{x^2-xy+y^2}\div(x^3-y^3)\) ?

Updated On: Aug 21, 2024
  • (x - y)/(x + y)
  • (x + y)/(x - y)
  • x + y
  • x - y
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The Correct Option is D

Solution and Explanation

We know, (a - b)2 = a2 - 2ab + b2
a4 + b4 + a2b2 = (a2 - ab + b2)(a2 + ab + b2)
a3 - b3 = (a - b)(a2 + ab - b2)
Then, \((x^4+y^4+x^2y^2)\frac{x^2-2xy+y^2}{x^2-xy+y^2}\div(x^3-y^3)\)
\(=(x^4+y^4+x^2y^2)\frac{x^2-2xy+y^2}{x^2-xy+y^2}(\frac{1}{x^3-y^3})\)
\(=(x^2-xy+y^2)(x^2+xy+y^2)\frac{(x-y)^2}{x^2-xy+y^2}(\frac{1}{(x-y)(x^2+xy+y^2)})\)
= x - y
So, the correct option is (D) : x - y.
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