If $a = b$ take $x=a$. If $a = -b$ take $x = a$ in the second equation. If $a = 0$, then take $x=0$ for the first equation. Suppose wlog that $|a| > |b|$. Let $Q(x) = x^4 - 2a^3x + b^4$ be the second polynomial. We see that $Q(a) = b^4 - a^4 < 0$ and $Q(-a) = 3a^4 + b^4 > 0$ and $0\neq a \neq -a$. So my the IVT $Q(x)$ has a real root.

first take a=b it shows (P(a)=0) P(x)=x^4-2a^3x+b^4, Q(x)=x^4-2b^3x+a^4) a don;t equal to b P(0) and Q(0) are big than 0 ((((( (P(a) = - Q(b) )))))) if they dont equal to 0 one of them will be smal than 0 after P(a) or Q(b) will be small than 0 after that then using polynomial is continious one of them will cross 0 in real number .

Let $f(x)=x^4-2b^3x+a^4$ and $g(x)=x^4-2a^3x+b^4.$ Assume the contrary, i.e. both equations do not have a real root. This means that $f(x)$ and $g(x)$ are positive for all real numbers $x$. Therefore, $$f\left(\frac{b}{\sqrt[3]{2}}\right)>0\iff a^4>\frac{3b^4}{2\sqrt[3]{2}}.$$Similarly, we have that $$g\left(\frac{a}{\sqrt[3]{2}}\right)>0\iff b^4>\frac{3a^4}{2\sqrt[3]{2}}.$$ Combining these two inequalities, we obtain that $$a^4>\frac{3b^4}{2\sqrt[3]{2}}>\frac{9a^4}{4\sqrt[3]{4}}\iff 4\sqrt[3]{4}>9,$$which is clearly false. Hence, our initial assumption was wrong and we are done.