(6-2ab-18^2b3)/12a⁴b

Question

0

871

1

Guest Aug 3, 2015

+118609

+5

Best Answer

Perhaps you mean:

$$\\\frac{(6-2ab-18^2b^3)}{12}\times a^4b\\\\
=\frac{(3-ab-9^2b^3)}{6}\times a^4b\\\\
=\frac{a^4b(3-ab-81b^3)}{6}\\\\$$

Now, if you want to, you could expand the brackets.

Melody Aug 4, 2015

Melody · Answer 1 · 2015-08-04T05:13:20

Perhaps you mean:

$$\\\frac{(6-2ab-18^2b^3)}{12}\times a^4b\\\\
=\frac{(3-ab-9^2b^3)}{6}\times a^4b\\\\
=\frac{a^4b(3-ab-81b^3)}{6}\\\\$$

Now, if you want to, you could expand the brackets.