## Algebra and Trigonometry 10th Edition

$4x^2(x-1)^\frac{1}{3}+6x(x-1)^\frac{4}{3}=(x-1)^\frac{1}{3}(4x^2+6x(x-1))=0$Hence either $(x-1)^\frac{1}{3}=0$, giving x=1 or $(4x^2+6x(x-1))=10x^2-6x=2x(5x-3)=0$ giving x=0 or 3/5