Find the following product:

Question:

Find the following product:

$\left(-\frac{7}{4} a b^{2} c-\frac{6}{25} a^{2} c^{2}\right)\left(-50 a^{2} b^{2} c^{2}\right)$

Solution:

To find the product, we will use distributive law as follows:

$\left(-\frac{7}{4} a b^{2} c-\frac{6}{25} a^{2} c^{2}\right)\left(-50 a^{2} b^{2} c^{2}\right)$

$=\left\{\left(-\frac{7}{4} a b^{2} c\right)\left(-50 a^{2} b^{2} c^{2}\right)\right\}-\left\{\left(\frac{6}{25} a^{2} c^{2}\right)\left(-50 a^{2} b^{2} c^{2}\right)\right\}$

$=\left\{\left\{-\frac{7}{4} \times(-50)\right\}\left(a \times a^{2}\right) \times\left(b^{2} \times b^{2}\right) \times\left(c \times c^{2}\right)\right\}-\left\{\left(\frac{6}{25}\right)(-50)\left(a^{2} \times a^{2}\right) \times\left(b^{2}\right) \times\left(c^{2} \times c^{2}\right)\right\}$

$=\left\{-\frac{7}{4} \times(-50)\right\}\left(a^{1+2} b^{2+2} c^{1+2}\right)-\left\{\left(\frac{6}{25}\right)(-50)\left(a^{2+2} b^{2} c^{2+2}\right)\right\}$

$=\frac{175}{2} a^{3} b^{4} c^{3}-\left(-12 a^{4} b^{2} c^{4}\right)$

$=\frac{175}{2} a^{3} b^{4} c^{3}+12 a^{4} b^{2} c^{4}$

Thus, the answer is $\frac{175}{2} a^{3} b^{4} c^{3}+12 a^{4} b^{2} c^{4}$.

 

 

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now