We expand this by substituting a = p, b = 3q and c = 7r into the identity. Squaring the individual items and finding their paired double products yields the answer.
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This expression matches the identity template a square – 2ab + b square. The middle variable terms cancel out completely when multiplying p/4 and 4/p, leaving only the constant integer minus 2.
This six-term polynomial matches the expanded three-term identity. The squared bases are m/3, k/2, and 3n, which merge into the single grouped factor (m/3 + k/2 + 3n) square.
This fits the identity a square + 2ab + b square. The first term is the square of 3/2 s and the last is the square of 2t, which yields the factor (3/2 s + 2t) square.
The polynomial matches the identity template a square – 2ab + b square. Since 16y square is (4y) square and 9 is (3) square, it condenses directly into (4y – 3) square.