We first take out the common constant factor 6 from the volume expression. This leaves a square – 4b square inside brackets, which breaks down further into (a + 2b)(a – 2b) using the difference of squares formula.
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We apply the difference of squares identity, x square – y square = (x + y)(x – y). Rewriting 36s square as (6s) square and 49t square as (7t) square directly provides the two unique linear dimensions.
The polynomial matches the identity template a square – 2ab + b square. Since 16s square is (4s) square and 25t square is (5t) square, it condenses directly into (4s – 5t) square.
We rewrite 97 as 100 – 3 and apply the subtraction square identity. Squaring 100 and 3, then subtracting twice their product gives the plain numerical solution.
We change 41 into 40 + 1 and apply the addition square identity. Squaring 40 and 1, then adding twice their product gives the plain numerical solution.