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.
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We treat 1120 as 1100 + 20 and apply the standard two-term addition square identity. Squaring both parts and adding their double product provides the final answer.
We express 198 as 200 – 2 and use the subtraction square identity with a = 200 and b = 2. Working out the arithmetic components gives the final plain text answer.
We express 117 as 100 + 10 + 7 and apply the three-term identity. Squaring each number and adding all the double-product combinations yields the correct numerical total.
Ayushree
Asked: In: Class 9 Maths
We transform 193 into 200 – 7 and apply the subtraction identity with a = 200 and b = 7. Evaluating these components together gives the correct final numerical answer of 37249.