A = \sqrt{35(35 - 17)(35 - 25)(35 - 28)} = \sqrt{35 \cdot 18 \cdot 10 \cdot 7} = \sqrt{44100} = 210

["Unlocking the Mystery: The Step-by-Step Calculation of A = √[35(35 − 17)(35 − 25)(35 − 28)] = 210", "Mathematics often hides elegant simplifications beneath complex expressions. One such intriguing example is the calculation of ( A = \sqrt{35(35 - 17)(35 - 25)(35 - 28)} ), which reveals a clean result of ( A = 210 ) in just a few logical steps. Whether you're a student aiming to master algebraic simplifications or a curious learner exploring algebraic identities, this problem showcases the power of factoring and simplification.", "---", "### Step-by-Step Breakdown of the Expression", "Start with the original expression:", "[\nA = \sqrt{35(35 - 17)(35 - 25)(35 - 28)}\n]", "### Step 1: Simplify the Inner Minuends", "Subtract each term inside the parentheses:", "[\n35 - 17 = 18\n]\n[\n35 - 25 = 10\n]\n[\n35 - 28 = 7\n]", "Substitute these simplified values:", "[\nA = \sqrt{35 \cdot 18 \cdot 10 \cdot 7}\n]", "---", "### Step 2: Rearrange the Factors for Easier Grouping", "Rearranging the terms creatively enables efficient multiplication and grouping:", "[\nA = \sqrt{35 \cdot 10 \cdot 18 \cdot 7}\n]", "Notice that ( 35 \cdot 7 = 245 ) and ( 10 \cdot 18 = 180 ). Now rewrite:", "[\nA = \sqrt{245 \cdot 180}\n]", "But to simplify even further—look instead for grouping that reveals perfect squares:", "[\n35 \cdot 10 = 350, \quad 18 \cdot 7 = 126 \quad \ ext{(too messy)}\n]", "Better yet: group as:", "[\n(35 \cdot 7) \cdot (10 \cdot 18) = 245 \cdot 180 \quad \ ext{(still not ideal)}\n]", "But let’s try pairing 35 and 10, and 18 and 7:", "[\n(35 \cdot 10) \cdot (18 \cdot 7) = 350 \cdot 126 — \ ext{not helpful}\n]", "Best grouping: Recognize that 35 = 5×7, and 10 = 2×5 — but let's instead reorder as:", "[\nA = \sqrt{35 \cdot 7 \cdot 10 \cdot 18}\n]", "Now group:", "[\n(35 \cdot 7) \cdot (10 \cdot 18) = 245 \cdot 180\n]", "Still not optimal. Try grouping 35 and 18, 10 and 7:", "[\n(35 \cdot 18) \cdot (10 \cdot 7) = 630 \cdot 70 = 44100\n]", "Still messy. Let’s try:", "Group as:", "[\nA = \sqrt{ (35 \cdot 10) \cdot (18 \cdot 7) } = \sqrt{350 \cdot 126}\n]", "No, better yet — observe:", "Try pairing for perfect squares directly:", "Let’s recompute numerically for insight:", "[\n35 - 17 = 18,\quad 35 - 25 = 10,\quad 35 - 28 = 7\n\Rightarrow \sqrt{35 \cdot 18 \cdot 10 \cdot 7}\n]", "Now factor each:", "- ( 35 = 5 \ imes 7 )\n- ( 18 = 2 \ imes 3^2 )\n- ( 10 = 2 \ imes 5 )\n- ( 7 = 7 )", "Now multiply all factors:", "[\n35 \cdot 18 \cdot 10 \cdot 7 = (5 \ imes 7) \cdot (2 \ imes 3^2) \cdot (2 \ imes 5) \cdot 7 = 2^2 \ imes 3^2 \ imes 5^2 \ imes 7^2\n]", "So the product is:", "[\n2^2 \ imes 3^2 \ imes 5^2 \ imes 7^2 = (2 \cdot 3 \cdot 5 \cdot 7)^2 = (210)^2\n]", "Therefore:", "[\nA = \sqrt{ (210)^2 } = 210\n]", "---", "### Why This Simplification Matters", "This example highlights how recognizing patterns — such as factor groups forming perfect squares — can reduce complex radicals to simple integers. Factoring before multplication reveals structure hidden in plain sight, turning a seemingly difficult equation into a straightforward squaring operation.", "---", "### Conclusion", "The equation\n[\nA = \sqrt{35(35 - 17)(35 - 25)(35 - 28)} = \sqrt{35 \cdot 18 \cdot 10 \cdot 7} = \sqrt{44100} = 210\n]\nis a perfect blend of subtraction, multiplication, and algebraic insight. Mastering such step-by-step simplifications not only boosts problem-solving confidence but also deepens understanding of number theory, radicals, and identities.", "Whether you're solving for math exams, tackling function values, or simply exploring algebra, this problem reminds us: sometimes the key to a complicated expression lies in careful simplification — and the joy of revealing ( A = 210 ), clean and clear.", "---", "### Bonus: Try It Yourself!", "Next time you see a radical with four linear terms like ( \sqrt{a(a - b)(a - c)(a - d)} ), remember this method:\n1. Simplify each binomial.\n2. Factor components into primes.\n3. Look for two squared pairs to form a perfect square.\n4. Conclude with the square root of the product — a perfect integer!", "This formulaic approach transforms complexity into clarity — and brings satisfaction with every square root resolved.", "---", "Key Takeaways:\n✔ Radicals can be simplified by careful factoring.\n✔ Recognizing perfect squares unlocks elegant solutions.\n✔ Algebraic intuition turns puzzles into proofs.\n✔ Practice with real numbers confirms patterns.", "Keywords:\n( \sqrt{35(35 - 17)(35 - 25)(35 - 28)} = 210 ),\nradical simplification, algebra, radical identity, factoring radicals, step-by-step math, perfect square under radical, multiplying binomials, math simplification."]