Question:

If A=x2−y2x2−y2A=\frac{x^2-y^2}{x^2-y^2}, B=x+yx2+y2+xyB=\frac{x+y}{x^2+y^2+xy} and C=3xy2−3x2yx2−y2C=\frac{ 3xy^2-3x^2y}{x^2-y^2}, then find the value of AB (A+C)

Updated On: Sep 13, 2024
  • (x−y)(x+y)\frac{(x-y)}{(x+y)}
  • (x−y)(x+y)2\frac{(x-y)}{(x+y)^2}
  • (x−y)2(x+y)\frac{(x-y)^2}{(x+y)}
  • (x−y)2(x+y)2\frac{(x-y)^2}{(x+y)^2}
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The Correct Option is C

Solution and Explanation

The correct option is (C): (x−y)2(x+y)\frac{(x-y)^2}{(x+y)}
AB(A+C)=x2−y3x2−y2×x+yx2+y2+xy(x3−x3x3−y2+3xy2−3x2yx2−y2)\frac{x^2-y^3}{x^2-y^2}\times\frac{x+y}{x^2+y^2+xy}(\frac{x^3-x^3}{x^3-y^2}+\frac{3xy^2-3x^2y}{x^2-y^2})
After simplyfying:
= (x−y)3(x+Y)(x−y))\frac{(x-y)^3}{(x+Y)(x-y))}
=(x−y)2(x+y)=\frac{(x-y)^2}{(x+y)}
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