2t^6 - 14t^3 + 2 = 0 \quad \Rightarrow \quad t^6 - 7t^3 + 1 = 0.

["Solving the Equation: (2t^6 - 14t^3 + 2 = 0) by Transforming to a Quadratic Form", "---", "### Introduction", "The equation (2t^6 - 14t^3 + 2 = 0) may appear complex at first glance, but with a clever substitution, it simplifies dramatically into a solvable quadratic. In this comprehensive guide, we explore how reducing this sixth-degree polynomial to a lower-degree form enables us to find all real and complex solutions efficiently. Understanding this transformation not only solves the equation but also demonstrates a powerful algebraic technique used in many areas of mathematics.", "This article explains step-by-step how to rewrite the original equation, solve the resulting quadratic, and finally interpret the roots. Whether you're a student tackling advanced algebra or a self-learner exploring substitution methods, this approach illuminates a key skill in polynomial solving.", "---", "### Rewriting the Equation: A Clever Substitution", "The original equation is:", "[\n2t^6 - 14t^3 + 2 = 0\n]", "Notice that (t^6 = (t^3)^2). This suggests that we can simplify the expression by substituting (u = t^3). Applying this substitution transforms the equation as follows:", "[\n2u^2 - 14u + 2 = 0\n]", "This is now a standard quadratic equation in variable (u), much easier to solve using well-known methods like factoring, completing the square, or the quadratic formula.", "---", "### Solving the Quadratic Equation", "We solve:", "[\n2u^2 - 14u + 2 = 0\n]", "Step 1: Simplify the equation\nDivide the entire equation by 2 to make coefficients smaller and easier:", "[\nu^2 - 7u + 1 = 0\n]", "Step 2: Apply the quadratic formula\nThe quadratic formula states that for (Au^2 + Bu + C = 0),\n[\nu = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\n]", "Here, (A = 1), (B = -7), (C = 1), so:", "[\nu = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(1)}}{2(1)} = \frac{7 \pm \sqrt{49 - 4}}{2} = \frac{7 \pm \sqrt{45}}{2}\n]", "Since (\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}), we have:", "[\nu = \frac{7 \pm 3\sqrt{5}}{2}\n]", "Thus, the two solutions for (u) are:", "[\nu_1 = \frac{7 + 3\sqrt{5}}{2}, \quad u_2 = \frac{7 - 3\sqrt{5}}{2}\n]", "---", "### Back-Substitution: Finding (t) from (t^3 = u)", "Recall that (u = t^3), so for each solution of (u), we solve (t^3 = u):", "1. (t^3 = \frac{7 + 3\sqrt{5}}{2})\n The real cube root is:\n [\n t = \sqrt[3]{\frac{7 + 3\sqrt{5}}{2}}\n ]\n There are also two complex cube roots, but these are all valid complex solutions.", "2. (t^3 = \frac{7 - 3\sqrt{5}}{2})\n Similarly:\n [\n t = \sqrt[3]{\frac{7 - 3\sqrt{5}}{2}}\n ]", "Each equation yields one real cube root and two complex roots (when considering multiplicity and the complex plane), but the principal real cube root per equation suffices in most practical applications.", "---", "### Summary of Solutions", "The equation (2t^6 - 14t^3 + 2 = 0) has six roots in total (since it's a sixth-degree polynomial), coming from the three cube roots for each of the two (u)-values. The solutions are:", "- Three cube roots from (t^3 = \frac{7 + 3\sqrt{5}}{2})\n- Three cube roots from (t^3 = \frac{7 - 3\sqrt{5}}{2})", "Explicitly:", "[\nt = \sqrt[3]{\frac{7 + 3\sqrt{5}}{2}},\quad\nt = \omega \sqrt[3]{\frac{7 + 3\sqrt{5}}{2}},\quad\nt = \omega^2 \sqrt[3]{\frac{7 + 3\sqrt{5}}{2}}\n]\n[\nt = \sqrt[3]{\frac{7 - 3\sqrt{5}}{2}},\quad\nt = \omega \sqrt[3]{\frac{7 - 3\sqrt{5}}{2}},\quad\nt = \omega^2 \sqrt[3]{\frac{7 - 3\sqrt{5}}{2}}\n]", "Here, (\omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i) is a primitive cube root of unity (solving (\omega^3 = 1), (\omega <br/>\ne 1)).", "---", "### Why This Substitution Works", "This method transforms a higher-degree polynomial into a quadratic, which is straightforward to solve. The key insight is recognizing (t^6 = (t^3)^2), allowing substitution. Such substitutions are common in algebra and famously used in solving equations like (z^4 - 5z^2 + 4 = 0), where (u = z^2).", "It also highlights the power of symmetry in polynomial equations—reducing complexity often reveals simpler underlying structures.", "---", "### Application and Implications", "Solving this equation has applications in fields involving polynomial modeling, signal processing, and control theory. More broadly, mastering substitution techniques like this strengthens algebraic reasoning and opens doors to solving other complex equations.", "---", "### Final Notes", "- The original equation (2t^6 - 14t^3 + 2 = 0) reduces to (u^2 - 7u + 1 = 0) via (u = t^3).\n- Two irrational real values for (u) yield real and complex (t) roots via cube roots.\n- All six roots stem from three cube roots applied to two values of (u).\n- This approach exemplifies elegance and efficiency in algebraic problem-solving.", "---", "### Conclusion", "To solve (2t^6 - 14t^3 + 2 = 0), substitute (u = t^3) to obtain a quadratic in (u), solve it with the quadratic formula, then take cube roots to recover (t). Although the equation appears daunting at first, clever substitution unlocks a clean pathway to all six roots—showcasing how transformation can simplify complexity.", "Whether preparing for exams, tackling research problems, or simply deepening your math skills, mastering this technique is invaluable.", "---", "Keywords: (2t^6 - 14t^3 + 2 = 0), solving polynomial equations, substitution method, quadratic form, cube roots, algebraic techniques, polynomial roots, complex solutions, algebraic manipulation.", "---", "Want more? Explore how this strategy extends to equations like (u^4 - 5u^2 + 4 = 0) or non-integer substitutions for deeper mastery."]