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(C) Given equation: $(0.64)^{4} \div (0.512)^{3} 	imes (0.8)^{4} = (0.8)^{?+3}$Express all terms with base $0.8$:$0.64 = (0.8)^{2}$$0.512 = (0.8)^{3}

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(C) Given equation: $(0.64)^{4} \div (0.512)^{3} 	imes (0.8)^{4} = (0.8)^{?+3}$Express all terms with base $0.8$:$0.64 = (0.8)^{2}$$0.512 = (0.8)^{3}$Substituting these values:$[(0.8)^{2}]^{4} \div [(0.8)^{3}]^{3} 	imes (0.8)^{4} = (0.8)^{?+3}$Using the power rule $(a^{m})^{n} = a^{m 	imes n}$:$(0.8)^{8} \div (0.8)^{9} 	imes (0.8)^{4} = (0.8)^{?+3}$Using the exponent rules $a^{m} \div a^{n} = a^{m-n}$ and $a^{m} 	imes a^{n} = a^{m+n}$:$(0.8)^{8-9+4} = (0.8)^{?+3}$$(0.8)^{3} = (0.8)^{?+3}$Equating the exponents:$3 = ? + 3$$? = 3 - 3 = 0$

What will come in the place of the question mark $(?)$ in the following equation?
$(0.64)^{4} \div (0.512)^{3} \times (0.8)^{4} = (0.8)^{?+3}$

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What will come in the place of the question mark $(?)$ in the following equation?$(0.64)^{4} \div (0.512)^{3} \times (0.8)^{4} = (0.8)^{?+3}$