[tex]\displaystyle
a+b=1 \\
a^3+b^3= \frac{1}{4} \\
\text{--------------------} \\
b=1-a \\
a^3+(1-a)^3 =\frac{1}{4} \\
a^3 + 1 -3a + 3a^2 -a^3=\frac{1}{4} \\
3a^2-3a+1=\frac{1}{4} ~~~~~| \times 4 \\
12a^2 - 12a + 4=1 \\
12a^2 - 12a + 3=0 ~~~~~|: 3 \\
4a^2 - 4a +1=0
[/tex]
[tex]\displaystyle
4a^2 - 4a +1=0 \\ \\
a_{_{12}}= \frac{4\pm \sqrt{16-4\times 4} }{8}= \frac{4\pm \sqrt{16-16} }{8} = \frac{4}{8} = \frac{1}{2} \\
a =\boxed{ \frac{1}{2} }~~~~~radacina~~dubla \\ \\
b= 1-a = 1-\frac{1}{2} = \boxed{ \frac{1}{2} }~~~~~radacina~~dubla \\ \\
\text{Solutia problemei:} \\ \\
\boxed{\boxed{a = b = \frac{1}{2} }}[/tex]
