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(B) Given the equations $a^{3} b = 180$ and $a b c = 180$.Since $a, b, c$ are positive integers,we equate the two expressions: $a^{3} b = a b c$.Divid

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$1$

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(B) Given the equations $a^{3} b = 180$ and $a b c = 180$.Since $a, b, c$ are positive integers,we equate the two expressions: $a^{3} b = a b c$.Dividing both sides by $ab$ (since $a, b > 0$),we get $a^{2} = c$.Now,we factorize $180$: $180 = 2^{2} 	imes 3^{2} 	imes 5 = 4 	imes 9 	imes 5$.We need to find $a$ such that $a^{2}$ is a factor of $180$ and $a^{3} b = 180$.If $a = 1$,then $c = a^{2} = 1^{2} = 1$. Then $1^{3} 	imes b = 180$,so $b = 180$. This satisfies the condition.If $a = 2$,then $c = a^{2} = 2^{2} = 4$. Then $2^{3} 	imes b = 180$,which gives $8b = 180$,so $b = 22.5$ (not an integer).If $a = 3$,then $c = a^{2} = 3^{2} = 9$. Then $3^{3} 	imes b = 180$,which gives $27b = 180$,so $b = 180/27 = 20/3$ (not an integer).Thus,the only integer solution is $a = 1, b = 180, c = 1$.

If $a^{3} b = a b c = 180$,and $a, b, c$ are positive integers,then the value of $c$ is

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