Using dimensional analysis on t=Aρarbηcσd:
the dimensions are [T] = [ML−3]a[L]b[ML−1T−1]c[ML−3]d.
Expanding: [T] = Ma+c+dL−3a+b−c−3dT−c.
Comparing powers: For T: 1=−c, giving c=−1.
For M: 0=a+c+d, so a+d=1.
For L: 0=−3a+b−c−3d=−3a+b+1−3d, giving b=3(a+d)−1=3(1)−1=2.
Therefore (b+c)/(a+d)=(2−1)/1=1.