(6-2ab-18^2b3)/12a⁴b

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Guest Aug 3, 2015

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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.