t[x_]:=(8(x^4)-8(x^2)+1)/8

p[x_]:=((x^2)-1) ((x^2)-(1/9))

Plot[{t[x],p[x]},{x,-1,1}]

Notice that the polynomial with evenly spaced zeros has significantly larger minima.  The blue polynomial has minima and maxima on the interval that are all the same distance from 0.

Solve[t[x]==0,x]

Using divided difference coeficients I got from Matlab, below I define the interpolating polynomials to (e^x) for the two different sets of nodes.  The first two plots show that both polynomials are awfully close to the function.

T[x_]:=0.396975968643480+(x+(1/2) (Sqrt[2+(Sqrt[2])]))(0.526708903907574+ (x+(1/2) (Sqrt[2-(Sqrt[2])])) (0.381059484997363+ (x-(1/2) (Sqrt[2-(Sqrt[2])]))0.175175694047241))

Plot[{(E^x),T[x], (E^x)-T[x]},{x,-1,1}]

V[x_]:=0.367879441171442+(x+1)(0.522977804103520+(x+(1/3)) (0.371732900748698+(x-(1/3))0.176151962109770))

Plot[{(E^x),V[x], (E^x)-V[x]},{x,-1,1}]

However, when we plot just the errors for the two, we can see that the equally spaced nodes definitely give a greater error.

Plot[{(E^x)-V[x], (E^x)-T[x]},{x,-1,1}]

c[x_]:=Piecewise[{{(x^(1/3)), x≥0},{-((-x)^(1/3)), x<0}}]

Plot[c[x],{x,-1,1}]