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(A) To solve the expression $\frac{343 	imes 49}{216 	imes 16 	imes 81}$,we express each number as a power of its prime factors:$343 = 7^{3}$$49 =

Vedclass · Accepted Answer

$\frac{7^{5}}{6^{7}}$

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(A) To solve the expression $\frac{343 	imes 49}{216 	imes 16 	imes 81}$,we express each number as a power of its prime factors:$343 = 7^{3}$$49 = 7^{2}$$216 = 6^{3}$$16 = 2^{4}$$81 = 3^{4}$Substituting these into the expression:$\frac{7^{3} 	imes 7^{2}}{6^{3} 	imes 2^{4} 	imes 3^{4}}$Since $2^{4} 	imes 3^{4} = (2 	imes 3)^{4} = 6^{4}$,the expression becomes:$\frac{7^{3+2}}{6^{3} 	imes 6^{4}} = \frac{7^{5}}{6^{3+4}} = \frac{7^{5}}{6^{7}}$

What will come in the place of the question mark $(?)$ in the following expression?
$\frac{343 \times 49}{216 \times 16 \times 81} = ?$

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What will come in the place of the question mark $(?)$ in the following expression?$\frac{343 \times 49}{216 \times 16 \times 81} = ?$