Solution: Let $ t = 2y + 1 \Rightarrow y = \frac{t - 1}{2} $. Substitute into $ h(t) $: $ h(t) = \left(\frac{t - 1}{2}\right)^3 - 5\left(\frac{t - 1}{2}\right) + 6 $. Simplify: $ h(t) = \frac{(t-1)^3}{8} - \frac{5(t-1)}{2} + 6 $. Now replace $ t $ with $ y^2 - 1 $: $ h(y^2 - 1) = \frac{(y^2 - 2)^3}{8} - \frac{5(y^2 - 2)}{2} + 6 $. Expand and combine terms (if required). \boxed{\frac{(y^2 - 2)^3}{8} - \frac{5(y^2 - 2)}{2} + 6}

["Simplifying Complex Functions: A Step-by-Step Solution", "Working with mathematical functions involving substitutions and expansions can seem daunting at first, but careful algebraic manipulation simplifies the process. In this article, we explore how substituting an expression into a function and simplifying step-by-step leads to a clearer, more elegant form—ideal for students, engineers, and scientists tackling advanced problem-solving.", "---", "### The Problem", "We begin with a function defined in terms of $ t $:\n$$\nh(t) = \left(\frac{t - 1}{2}\right)^3 - 5\left(\frac{t - 1}{2}\right) + 6\n$$", "The goal is to substitute $ t = 2y + 1 $, simplify the expression, and ultimately replace $ t $ with $ y^2 - 1 $ to obtain $ h(y^2 - 1) $ in its simplest expanded form.", "---", "### Step 1: Substitution", "Given $ t = 2y + 1 $, substitute into $ h(t) $:\n$$\nh(2y + 1) = \left(\frac{(2y + 1) - 1}{2}\right)^3 - 5\left(\frac{(2y + 1) - 1}{2}\right) + 6\n$$\nSimplify the expression inside the parentheses:\n$$\n\frac{2y}{2} = y\n$$\nSo,\n$$\nh(2y + 1) = \left(\frac{t - 1}{2}\right)^3 \rightarrow y^3 \quad \ ext{when substituted}\n$$\nBut let’s proceed carefully:\n$$\nh(t) = \left(\frac{t - 1}{2}\right)^3 - \frac{5(t - 1)}{2} + 6\n\quad \Rightarrow \quad h(2y + 1) = \left(\frac{2y}{2}\right)^3 - \frac{5(2y)}{2} + 6 = y^3 - 5y + 6\n$$", "---", "### Step 2: Express in Terms of $ y^2 - 1 $", "Now, we replace $ t $ with $ y^2 - 1 $:\n$$\nh(y^2 - 1) = \left(\frac{(y^2 - 1) - 1}{2}\right)^3 - 5\left(\frac{(y^2 - 1) - 1}{2}\right) + 6\n$$\nSimplify:\n$$\n\frac{(y^2 - 2)}{2}\n\Rightarrow\nh(y^2 - 1) = \left(\frac{y^2 - 2}{2}\right)^3 - 5\left(\frac{y^2 - 2}{2}\right) + 6\n$$", "---", "### Step 3: Expand and Combine Terms", "Now expand each term.", "First term:\n$$\n\left(\frac{y^2 - 2}{2}\right)^3 = \frac{(y^2 - 2)^3}{8}\n$$\nExpand $ (y^2 - 2)^3 $ using the binomial theorem:\n$$\n(y^2 - 2)^3 = y^6 - 3 \cdot y^4 \cdot 2 + 3 \cdot y^2 \cdot 4 - 8 = y^6 - 6y^4 + 12y^2 - 8\n$$\nSo:\n$$\n\frac{(y^2 - 2)^3}{8} = \frac{y^6 - 6y^4 + 12y^2 - 8}{8}\n$$", "Second term:\n$$\n-5 \cdot \frac{y^2 - 2}{2} = -\frac{5y^2 - 10}{2}\n= -\frac{5y^2}{2} + 5\n$$", "Third term: $ +6 $", "---", "### Combine all terms:", "$$\nh(y^2 - 1) = \frac{y^6 - 6y^4 + 12y^2 - 8}{8} - \frac{5y^2}{2} + 5 + 6\n$$\nConvert constants and fractions to eighths:\n$$\n5 + 6 = 11 = \frac{88}{8}, \quad -\frac{5y^2}{2} = -\frac{20y^2}{8}, \quad -\frac{8}{8} \ ext{ from numerator}\n$$\nSo:\n$$\nh(y^2 - 1) = \frac{y^6 - 6y^4 + 12y^2 - 8 - 20y^2 + 88}{8}\n= \frac{y^6 - 6y^4 - 8y^2 + 80}{8}\n$$", "---", "### Final Simplified Form", "$$\n\boxed{h(y^2 - 1) = \frac{(y^2 - 2)^3}{8} - \frac{5(y^2 - 2)}{2} + 6 = \frac{y^6 - 6y^4 - 8y^2 + 80}{8}}\n$$", "---", "### Why This Matters", "This type of algebraic simplification supports numerous applications in engineering, signal processing, and computer graphics—where substitution and expansion transform complex models into manageable forms. Mastering these steps improves problem clarity and computational accuracy.", "---", "Key takeaway: Breaking down a function step-by-step—substitution, expansion, and simplification—turns complexity into comprehension. Whether in math or applied fields, clean mathematical expression is power."]