Answer:
[tex]x_1=\dfrac{a(a^2+ab+b^2)}{(a+b)^2}\\ \\x_2=\dfrac{b(a^2+ab+b^2)}{(a+b)^2}[/tex]
Step-by-step explanation:
First, simplify the left part:
[tex]\dfrac{a}{a-x}+\dfrac{b}{b-x}=\dfrac{a(b-x)+b(a-x)}{(a-x)(b-x)}=\dfrac{ab-ax+ab-bx}{ab-ax-bx+x^2}[/tex]
Now, cross multiply:
[tex]ab\cdot (ab-ax+ab-bx)=(a+b)^2\cdot (ab-ax-bx+x^2)\\ \\2(ab)^2-a^2bx-ab^2x=(a^2+2ab+b^2)(ab-ax-bx+x^2)\\ \\2(ab)^2-a^2bx-ab^2x=a^3b-a^3x-a^2bx+a^2x^2+2(ab)^2-2a^2bx-2ab^2x+2abx^2+ab^3-ab^2x-b^3x+b^2x^2\\ \\a^3b-a^3x+a^2x^2-2a^2bx-2ab^2x+2abx^2+ab^3-b^3x+b^2x^2=0\\ \\x^2(a^2+2ab+b^2)+x(-a^3-2a^2b-2ab^2-b^3)+a^3b+ab^3=0\\ \\(a+b)^2x^2-x((a+b)(a^2-ab+b^2)+2ab(a+b))+ab(a^2+b^2)=0\\ \\(a+b)^2x^2-x((a+b)(a^2+ab+b^2))+ab(a^2+b^2)=0\\ \\(a+b)^2x^2-x(a+b)(a^2+ab+b^2)+ab(a^2+b^2)=0[/tex]
[tex]D=(a+b)^2(a^2+ab+b^2)^2-4(a+b)^2ab(a^2+b^2)\\ \\=(a+b)^2(a^4+(ab)^2+b^4+2a^3b+2a^2b^2+2ab^3-4a^3b-4ab^3)\\ \\=(a+b)^2(a^2-ab+b^2)^2\\ \\\sqrt{D}=(a+b)(a^2-ab+b^2)=a^3-b^3[/tex]
So,
[tex]x_{1,2}=\dfrac{(a+b)(a^2+ab+b^2)\pm (a^3-b^3)}{2(a+b)^2}\\ \\x_1=\dfrac{a^3+a^2b+ab^2+ba^2+ab^2+b^3+a^3-b^3}{2(a+b)^2}=\dfrac{2a^3+2a^2b+2ab^2}{2(a+b)^2}=\dfrac{2a(a^2+ab+b^2)}{2(a+b)^2}=\dfrac{a(a^2+ab+b^2)}{(a+b)^2}\\ \\x_2=\dfrac{a^3+a^2b+ab^2+ba^2+ab^2+b^3-a^3+b^3}{2(a+b)^2}=\dfrac{2b^3+2a^2b+2ab^2}{2(a+b)^2}=\dfrac{2b(a^2+ab+b^2)}{2(a+b)^2}=\dfrac{b(a^2+ab+b^2)}{(a+b)^2}[/tex]
