Recall perfect squares to 25, common cubes, and their roots forward and backward.
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Squares: \(11^2 = 121\), \(12^2 = 144\), \(13^2 = 169\), \(14^2 = 196\), \(15^2 = 225\), \(16^2 = 256\), \(17^2 = 289\), \(18^2 = 324\), \(19^2 = 361\), \(20^2 = 400\), \(21^2 = 441\), \(24^2 = 576\), \(25^2 = 625\).
Cubes: \(2^3 = 8\), \(3^3 = 27\), \(4^3 = 64\), \(5^3 = 125\), \(6^3 = 216\), \(7^3 = 343\), \(8^3 = 512\), \(9^3 = 729\), \(10^3 = 1000\).
Roots work backward: \(\sqrt{441} = 21\) because \(21^2 = 441\). The last digit helps: a square that ends in 1 comes from a number that ends in 1 or 9.
Right-triangle triples save time: 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29, 9-40-41, and their multiples (6-8-10, 10-24-26).
Common slips: stopping at \(a^2 + b^2\) and forgetting the square root, and mixing up \(5^3 = 125\) with \(5 \times 3 = 15\).
By the Pythagorean theorem, \(c = \sqrt{12^2 + 35^2} = \sqrt{1369} = 37\) in. (Stopping at \(12^2 + 35^2 = 1369\) forgets the square root. Adding the legs, \(12 + 35\), gives 47. Subtracting the squares, \(\sqrt{35^2 - 12^2}\), gives \(\sqrt{1081}\).)
By the Pythagorean theorem, the other leg \(= \sqrt{15^2 - 12^2} = \sqrt{81} = 9\) m. (Stopping at \(15^2 - 12^2 = 81\) forgets the square root. Adding the squares, as if 15 were a leg, gives \(3\sqrt{41}\). Subtracting the lengths without squaring, \(15 - 12\), gives 3.)
Factor: \(150 = 2 \times 3 \times 5^{2}\). In a perfect square every exponent is even. The primes with odd exponents are \(2\), \(3\), so multiply by one more of each: \(m = 6\). Check: \(150 \times 6 = 900 = 30^{2}\). Fixing only the exponent of 2 gives \(2\). Fixing only the exponent of 3 gives \(3\). Making a perfect cube instead of a perfect square gives \(180\).