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(C) To evaluate $103 	imes 105$ without direct multiplication,we can use the algebraic identity: $(x + a)(x + b) = x^2 + (a + b)x + ab$.Here,we can w

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$10815$

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(C) To evaluate $103 	imes 105$ without direct multiplication,we can use the algebraic identity: $(x + a)(x + b) = x^2 + (a + b)x + ab$.Here,we can write $103 = (100 + 3)$ and $105 = (100 + 5)$.Comparing this with the identity,we have $x = 100$,$a = 3$,and $b = 5$.Substituting these values into the identity:$(100 + 3)(100 + 5) = (100)^2 + (3 + 5)(100) + (3 	imes 5)$$= 10000 + (8 	imes 100) + 15$$= 10000 + 800 + 15$$= 10815$.

Evaluate the following product without multiplying directly: $103 \times 105$.

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