Question
Find the product: $\big(−\frac{7}{4}\text{ab}^2\text{c}−\frac{6}{25}\text{a}^2\text{c}^2\big)\big(−50\text{a}^2\text{b}^2\text{c}^2\big)$
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Answer
To find the product, we will use distributive law as follows:$\big(−\frac{7}{4}\text{ab}^2\text{c}−\frac{6}{25}\text{a}^2\text{c}^2\big)\big(−50\text{a}^2\text{b}^2\text{c}^2\big)$
$=\Big\{\big(−\frac{7}{4}\text{ab}^2\text{c}\big)\big(−50\text{a}^2\text{b}^2\text{c}^2\big)\Big\}−\Big\{\big(\frac{6}{25}\text{a}^2\text{c}^2\big)(−50\text{a}^2\text{b}^2\text{c}^2\big)\Big\}$
$=\Big\{−\frac{7}{4}×(−50)\Big\}\big(\text{a}×\text{a}^2\big)×\big(\text{b}^2×\text{b}^2\big)×\big(\text{c}×\text{c}^2\big)\Big\}\\-\Big\{\big(\frac{6}{25}\big)(−50)\Big\}\big(\text{a}^2×\text{a}^2\big)×\big(\text{b}^2\big)×\big(\text{c}^2×\text{c}^2\big)\Big\}$
$=\Big\{−\frac{7}{4}×(−50)\Big\}\big(\text{a}^{1+2}\text{b}^{2+2}\text{c}^{1+2}\big)\\-\Big\{\big(\frac{6}{25}\big)(−50)\big(\text{a}^2+2\text{b}^2\text{c}^{2+2}\Big\}$
$=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3−\big(−12\text{a}^4\text{b}^2\text{c}^4\big)$
$=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3+12\text{a}^4\text{b}^2\text{c}^4$
Thus, the answer is $=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3+12\text{a}^4\text{b}^2\text{c}^4$
$=\Big\{\big(−\frac{7}{4}\text{ab}^2\text{c}\big)\big(−50\text{a}^2\text{b}^2\text{c}^2\big)\Big\}−\Big\{\big(\frac{6}{25}\text{a}^2\text{c}^2\big)(−50\text{a}^2\text{b}^2\text{c}^2\big)\Big\}$
$=\Big\{−\frac{7}{4}×(−50)\Big\}\big(\text{a}×\text{a}^2\big)×\big(\text{b}^2×\text{b}^2\big)×\big(\text{c}×\text{c}^2\big)\Big\}\\-\Big\{\big(\frac{6}{25}\big)(−50)\Big\}\big(\text{a}^2×\text{a}^2\big)×\big(\text{b}^2\big)×\big(\text{c}^2×\text{c}^2\big)\Big\}$
$=\Big\{−\frac{7}{4}×(−50)\Big\}\big(\text{a}^{1+2}\text{b}^{2+2}\text{c}^{1+2}\big)\\-\Big\{\big(\frac{6}{25}\big)(−50)\big(\text{a}^2+2\text{b}^2\text{c}^{2+2}\Big\}$
$=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3−\big(−12\text{a}^4\text{b}^2\text{c}^4\big)$
$=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3+12\text{a}^4\text{b}^2\text{c}^4$
Thus, the answer is $=\frac{175}{2}\text{a}^3\text{b}^4\text{c}^3+12\text{a}^4\text{b}^2\text{c}^4$
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