Solution: Let $ v = \sqrt{u} $, so $ u = v^2 $. Substituting into the equation gives $ v^3 - 4v^2 + 3v = 0 $. Factoring yields $ v(v^2 - 4v + 3) = 0 $, which simplifies to $ v(v-1)(v-3) = 0 $. Thus, $ v = 0, 1, 3 $, corresponding to $ u = 0, 1, 9 $. The sum of the roots is $ 0 + 1 + 9 = 10 $. oxed{10}

["Solving Quadratic-Type Equations Using Substitution: A Step-by-Step Example", "Mathematics often presents complex equations that can be simplified through clever substitutions. One insightful method involves letting ( v = \sqrt{u} ), which transforms an equation involving square roots into a polynomial form—easier to analyze and solve. In this article, we explore this technique using the equation ( v^3 - 4v^2 + 3v = 0 ), demonstrating how substitution leads to simple factoring and reveals all solutions clearly.", "---", "### The Original Equation", "Consider the equation:\n[\nv^3 - 4v^2 + 3v = 0\n]\nAt first glance, it appears nonlinear and potentially difficult to solve directly. However, by recognizing ( v = \sqrt{u} ), we exploit the relationship ( u = v^2 ) to uncover a cubic polynomial in ( v ), simplifying our approach.", "---", "### Substitution to Reduce Degree", "From ( v = \sqrt{u} ), squaring both sides gives ( u = v^2 ). Substituting ( u = v^2 ) into the original equation yields:\n[\nv^3 - 4v^2 + 3v = 0\n]\nThis transformation converts a cubic in ( v ) into a manageable cubic expression.", "---", "### Factoring the Polynomial", "Factor out ( v ):\n[\nv(v^2 - 4v + 3) = 0\n]\nNow factor the quadratic:\n[\nv(v - 1)(v - 3) = 0\n]", "Setting each factor to zero gives the solutions:\n[\nv = 0,\quad v = 1,\quad v = 3\n]", "---", "### Recovering Solutions in Terms of ( u )", "Since ( u = v^2 ), substitute each value of ( v ):\n- If ( v = 0 ), then ( u = 0^2 = 0 )\n- If ( v = 1 ), then ( u = 1^2 = 1 )\n- If ( v = 3 ), then ( u = 3^2 = 9 )", "Thus, the solutions in ( u ) are ( u = 0,\ 1,\ 9 ).", "---", "### Sum of the Roots", "The sum of the roots in ( u ) is:\n[\n0 + 1 + 9 = 10\n]", "This result underscores how substitution not only simplifies solving but also reveals full root structure through algebraic factoring.", "---", "### Why This Method Works", "Using substitution like ( v = \sqrt{u} ) is powerful because:\n- It converts nonlinear equations into polynomials.\n- It enables factoring techniques familiar in algebra.\n- It simplifies complex root-finding tasks, especially for equations involving roots or radicals.", "---", "### Final Takeaway", "This example highlights a key strategy in solving equations with radicals: substitution transforms complexity into clarity. Recognizing patterns, factoring, and translating back to original variables ensures accurate and efficient solutions.", "So, next time you face an equation like ( v^3 - 4v^2 + 3v = 0 ), try substituting ( v = \sqrt{u} )—your solution path may become much simpler.", "---", "(\boxed{10})"]