#### 627,200Question: Let $ a $ and $ b $ be positive real numbers such that $ \frac{a}{b} + \frac{b}{a} = 5 $. Find the value of $ \frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2} $.

["Title: Compute $ \frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2} $ Given $ \frac{a}{b} + \frac{b}{a} = 5 $", "---", "Meta Description:\nGiven $ \frac{a}{b} + \frac{b}{a} = 5 $, find the exact value of $ \frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2} $ using algebraic identities and substitution.", "---", "### Introduction", "We are given that $ a $ and $ b $ are positive real numbers satisfying:", "$$\n\frac{a}{b} + \frac{b}{a} = 5\n$$", "Our goal is to compute:", "$$\nE = \frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2}\n$$", "This expression combines symmetric rational functions of $ a $ and $ b $, reminiscent of the original fraction condition. Let’s simplify step by step using clever substitutions and identities.", "---", "### Step 1: Simplify using substitution", "Let $ x = \frac{a}{b} $. Since $ a, b > 0 $, $ x > 0 $. Then:", "$$\n\frac{a}{b} + \frac{b}{a} = x + \frac{1}{x} = 5\n$$", "Multiply both sides by $ x $:", "$$\nx^2 + 1 = 5x \Rightarrow x^2 - 5x + 1 = 0\n$$", "This quadratic equation won’t be needed directly, but it helps us express symmetric functions in terms of $ x $.", "Note that:", "- $ \frac{a^2}{b^2} = x^2 $, so $ a^2 = x^2 b^2 $\n- We can factor $ a^2, b^2 $ in numerator and denominator expressions.", "Now express $ E $ in terms of $ x $:", "$$\nE = \frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2}\n$$", "Divide numerator and denominator in each term by $ b^2 $ (since $ b > 0 $):", "Let $ x^2 = \frac{a^2}{b^2} $. Then:", "$$\n\frac{a^2 + b^2}{a^2 - b^2} = \frac{x^2 + 1}{x^2 - 1}, \quad \frac{a^2 - b^2}{a^2 + b^2} = \frac{x^2 - 1}{x^2 + 1}\n$$", "So:", "$$\nE = \frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1}\n$$", "Let $ u = \frac{x^2 + 1}{x^2 - 1} $, then $ \frac{x^2 - 1}{x^2 + 1} = \frac{1}{u} $, so:", "$$\nE = u + \frac{1}{u}, \quad \ ext{where } u = \frac{x^2 + 1}{x^2 - 1}\n$$", "We now compute $ E = u + \frac{1}{u} = \frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1} $", "---", "### Step 2: Use identity to simplify $ E = u + \frac{1}{u} $", "Let’s compute $ E $ directly:", "$$\nE = \frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1}\n= \frac{(x^2 + 1)^2 + (x^2 - 1)^2}{(x^2 - 1)(x^2 + 1)}\n$$", "Expand numerator:", "$$\n(x^2 + 1)^2 = x^4 + 2x^2 + 1 \\n(x^2 - 1)^2 = x^4 - 2x^2 + 1 \\n\ ext{Sum} = (x^4 + 2x^2 + 1) + (x^4 - 2x^2 + 1) = 2x^4 + 2\n$$", "Denominator:", "$$\n(x^2 - 1)(x^2 + 1) = x^4 - 1\n$$", "So:", "$$\nE = \frac{2x^4 + 2}{x^4 - 1} = \frac{2(x^4 + 1)}{x^4 - 1}\n$$", "Now, recall from earlier: $ x + \frac{1}{x} = 5 $. Square both sides:", "$$\n\left(x + \frac{1}{x}\right)^2 = 25 \Rightarrow x^2 + 2 + \frac{1}{x^2} = 25 \Rightarrow x^2 + \frac{1}{x^2} = 23\n$$", "Now compute $ x^4 + \frac{1}{x^4} $. Square again:", "$$\n\left(x^2 + \frac{1}{x^2}\right)^2 = 23^2 = 529 \Rightarrow x^4 + 2 + \frac{1}{x^4} = 529 \Rightarrow x^4 + \frac{1}{x^4} = 527\n$$", "Now divide numerator and denominator of $ E = \frac{2(x^4 + 1)}{x^4 - 1} $ by $ x^2 $ to bring it into terms of $ x^2 + \frac{1}{x^2} $, but instead, observe:", "Let’s write:", "$$\nE = \frac{2(x^4 + 1)}{x^4 - 1} = 2 \cdot \frac{x^4 + 1}{x^4 - 1}\n$$", "Now write $ x^4 + 1 = (x^4 - 1) + 2 $, so:", "$$\nE = 2 \cdot \frac{(x^4 - 1) + 2}{x^4 - 1} = 2\left(1 + \frac{2}{x^4 - 1}\right) = 2 + \frac{4}{x^4 - 1}\n$$", "But this still depends on $ x^4 $, so we need a better path.", "Instead, note that $ E = \frac{2(x^4 + 1)}{x^4 - 1} $, and we know $ x^4 + \frac{1}{x^4} = 527 $. Let’s denote $ y = x^4 $, so $ y + \frac{1}{y} = 527 $. Multiply both sides by $ y $: $ y^2 - 527y + 1 = 0 $. This is complicated, but we don’t need $ y $ explicitly.", "Note:", "We want $ E = \frac{2(y + 1)}{y - 1} $, since $ y = x^4 $", "Let’s compute $ \frac{2(y + 1)}{y - 1} $. Write:", "$$\n\frac{2(y + 1)}{y - 1} = 2 \cdot \frac{y + 1}{y - 1}\n$$", "Now consider:", "$$\n\frac{y + 1}{y - 1} = \frac{(y - 1) + 2}{y - 1} = 1 + \frac{2}{y - 1}\n$$", "Still not helpful. But note: we already know $ x + \frac{1}{x} = 5 $, and from that we derived $ x^2 + \frac{1}{x^2} = 23 $, $ x^4 + \frac{1}{x^4} = 527 $. But our expression is in $ x^4 $, not reciprocal.", "Wait — revisit expression:", "We had:", "$$\nE = \frac{2x^4 + 2}{x^4 - 1} = 2 \cdot \frac{x^4 + 1}{x^4 - 1}\n$$", "Let’s compute $ \frac{x^4 + 1}{x^4 - 1} $ using known identities.", "But here’s a key insight: let’s go back to earlier:", "We had $ u = \frac{x^2 + 1}{x^2 - 1} $, and $ x + \frac{1}{x} = 5 $. Can we compute $ u $ directly?", "Let $ x + \frac{1}{x} = 5 $. Let’s compute $ x^2 + \frac{1}{x^2} = 23 $, as before.", "Now compute $ x^2 + \frac{1}{x^2} = 23 \Rightarrow \frac{x^4 + 1}{x^2} = 23 \Rightarrow x^4 + 1 = 23x^2 $", "Similarly, $ x^4 - 1 = (x^2 - 1)(x^2 + 1) $, but let’s use:", "From $ x^4 + 1 = 23x^2 $, so $ x^4 + 1 = 23x^2 $", "But earlier numerator in $ E $ was $ 2x^4 + 2 = 2(x^4 + 1) = 2 \cdot 23x^2 = 46x^2 $", "Denominator: $ x^4 - 1 = (x^2 - 1)(x^2 + 1) $", "But $ x^4 - 1 = x^4 + 1 - 2 = 23x^2 - 2 $? Not helpful.", "Wait — use $ x^4 + 1 = 23x^2 \Rightarrow x^4 - 1 = 23x^2 - 2 $? Still messy.", "Alternative idea: go back to $ E = \frac{2(x^4 + 1)}{x^4 - 1} $, and express $ x^4 + 1 $ and $ x^4 - 1 $ in terms of powers.", "But here’s a better idea: define $ z = x^2 $. Then $ z + \frac{1}{z} = x^2 + \frac{1}{x^2} = 23 $", "Let $ z + \frac{1}{z} = 23 $. We want:", "$$\nE = \frac{2(z^2 + 1)}{z^2 - 1}\n$$", "Now compute $ z^2 + \frac{1}{z^2} = \left(z + \frac{1}{z}\right)^2 - 2 = 23^2 - 2 = 529 - 2 = 527 $", "But again, we need $ \frac{2(z^2 + 1)}{z^2 - 1} $", "Wait — $ \frac{2(z^2 + 1)}{z^2 - 1} = 2 \cdot \frac{z^2 + 1}{z^2 - 1} $", "Let’s compute $ \frac{z^2 + 1}{z^2 - 1} $. This depends only on $ z + \frac{1}{z} $, but not directly.", "Instead, define $ w = z - \frac{1}{z} $, but perhaps compute numerically? No — must be exact.", "Wait — go back to expression:", "We had $ E = \frac{2x^4 + 2}{x^4 - 1} $, and we know $ x^4 + \frac{1}{x^4} = 527 $. Let $ y = x^4 $. Then:", "$$\nE = \frac{2y + 2}{y - 1}\n$$", "Multiply numerator and denominator by $ x^4 $:", "But better: write:", "$$\nE = \frac{2y + 2}{y - 1} = 2 \cdot \frac{y + 1}{y - 1}\n$$", "Now recall from $ x + \frac{1}{x} = 5 $, we derived $ x^2 + \frac{1}{x^2} = 23 $, and $ x^4 + \frac{1}{x^4} = 527 $. But we need a relation involving $ x^4 $, not $ 1/x^4 $.", "But observe: $ \frac{x^4 + 1}{x^4 - 1} = \frac{y + 1}{y - 1} $. Let’s express this using $ s = x + \frac{1}{x} = 5 $, and known symmetric sums.", "But here’s a breakthrough: let’s compute $ E = \frac{2(x^4 + 1)}{x^4 - 1} $, and use:", "Let’s divide numerator and denominator by $ x^2 $:", "$$\n\frac{2(x^4 + 1)}{x^4 - 1} = 2 \cdot \frac{x^2 + \frac{1}{x^2}}{x^2 - \frac{1}{x^2}}\n$$", "Because divide numerator and denominator by $ x^2 $:", "Numerator: $ 2(x^4 + 1)/x^2 = 2(x^2 + 1/x^2) $", "Denominator: $ (x^4 - 1)/x^2 = x^2 - 1/x^2 $", "Yes! So:", "$$\nE = 2 \cdot \frac{x^2 + \frac{"]