What is (12a^3b^2c-18a^4b^5c^2)/(6a^2bc) ?

The solution to (12a^3b^2c-18a^4b^5c^2)/(6a^2bc) is ab(2-3ab^3c)

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simplify (12a^3b^2c-18a^4b^5c^2)/(6a^2bc)

simplify

\frac{12 a ^{3} b ^{2} c - 18 a ^{4} b ^{5} c ^{2}}{6 a ^{2} b c}

ab (2 - 3 a b^{3} c)

Solution steps

\frac{12 a ^{3} b ^{2} c - 18 a ^{4} b ^{5} c ^{2}}{6 a ^{2} b c}

Factor out common term

6 a^{3} b^{2} c : 6 a^{3} b^{2} c (2 - 3 a b^{3} c)

= \frac{6 a ^{3} b ^{2} c ( 2 - 3 a b ^{3} c )}{6 a ^{2} b c}

Cancel the common factor:

6

= \frac{a ^{3} b ^{2} c ( 2 - 3 a b ^{3} c )}{a ^{2} b c}

Cancel the common factor:

c

= \frac{a ^{3} b ^{2} ( 2 - 3 a b ^{3} c )}{a ^{2} b}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

a^{3} = a^{2} a

= \frac{a ^{2} a b ^{2} ( 2 - 3 a b ^{3} c )}{a ^{2} b}

Cancel the common factor:

a^{2}

= \frac{a b ^{2} ( 2 - 3 a b ^{3} c )}{b}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

b^{2} = bb

= \frac{abb ( 2 - 3 a b ^{3} c )}{b}

Cancel the common factor:

b

= ab (2 - 3 a b^{3} c)

2^{- 0.5}

8 (x - 2)

in (1)

(8 - x^{2})^{2}

4 x^{2} - 6

What is (12a^3b^2c-18a^4b^5c^2)/(6a^2bc) ?
The solution to (12a^3b^2c-18a^4b^5c^2)/(6a^2bc) is ab(2-3ab^3c)