After expansion: $$(a+b)x^2-(2a^2+2b^2)x+a^3+b^3=0$$has a real solution. Thus it is of the form $(a+b)(x-r)^2=0$ where $r$ is the unique root. Comparing coefficients gives: $$a^2+b^2=r(a+b),a^3+b^3=r^2(a+b)$$so $r=\frac{a^2+b^2}{a+b}=\frac{a^3+b^3}{a^2+b^2}$ (we can assume that $a+b\ne0$, since the alternative is $a=\pm b$). But: $$0=\frac{a^3+b^3}{a^2+b^2}-\frac{a^2+b^2}{a+b}=\frac{ab}{(a^2+b^2)(a+b)}(a-b)^2$$and $\frac{ab}{(a^2+b^2)(a+b)}\ne0$, so we have to have $a=b$.