Question

given, B=X^2+XY+Y^2 And D=X^4+X^2 Y^2+Y^4 D÷B = ?
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Answer 1

Answer:

The answer is $x^2-xy+y^2$

Step-by-step explanation:

Given

        $B=x^2+xy+y^2$

And $D=x^4+x^2y^2+y^4$

            $=(x^2)^2+2(x^2)(y^2)+(y^2)^2-(x^2y^2)$, add and subtract $x^2y^2$

           $=(x^2+y^2)^2-(xy)^2\\=(x^2+y^2+xy)(x^2+y^2-xy)$

Therefore

      $\dfrac{D}{B}=\dfrac{(x^2+xy+y^2)(x^2-xy+y^2)}{x^2+xy+y^2}=x^2-xy+y^2$