We separate 1104 into 1000 + 100 + 4 and expand it using the three-term identity. Adding the squares and all two-term product pairings gives the total value.
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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 rewrite 79 as 80 – 1 to use the subtraction square identity with a = 80 and b = 1. Combining their squared values and subtracting the double product gives 6241.
Following the hint in the book, we factor out 1/3 as a common factor. This leaves the expression 9a square + 12ab + 4b square inside brackets, which factors into (3a + 2b) square.
This polynomial fits the standard identity a square + 2ab + b square. We rewrite 4s square as (2s) square and 25t square as (5t) square, giving the factored result of (2s + 5t) square.