What is (9a^4b^3-3a^3b^4)/(6a^3b^3-2a^2b^4) ?

The solution to (9a^4b^3-3a^3b^4)/(6a^3b^3-2a^2b^4) is (3a)/2

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simplify (9a^4b^3-3a^3b^4)/(6a^3b^3-2a^2b^4)

simplify

\frac{9 a ^{4} b ^{3} - 3 a ^{3} b ^{4}}{6 a ^{3} b ^{3} - 2 a ^{2} b ^{4}}

\frac{3 a}{2}

Solution steps

\frac{9 a ^{4} b ^{3} - 3 a ^{3} b ^{4}}{6 a ^{3} b ^{3} - 2 a ^{2} b ^{4}}

Factor

9 a^{4} b^{3} - 3 a^{3} b^{4} : 3 a^{3} b^{3} (3 a - b)

= \frac{3 a ^{3} b ^{3} ( 3 a - b )}{6 a ^{3} b ^{3} - 2 a ^{2} b ^{4}}

Factor

6 a^{3} b^{3} - 2 a^{2} b^{4} : 2 a^{2} b^{3} (3 a - b)

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

Cancel the common factor:

b^{3}

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

Cancel the common factor:

3 a - b

= \frac{3 a ^{3}}{2 a ^{2}}

Apply exponent rule:

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

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

= \frac{3 a ^{2} a}{2 a ^{2}}

Cancel the common factor:

a^{2}

= \frac{3 a}{2}

\frac{2}{x + 2} > \frac{1}{x + 3}

2 (3 x - 4) \leq - 20

- 16 x^{2} - 6 x^{4}

- 3 (5 y - 4) \geq 17

y^{2} - 3 y + 9 = 0

What is (9a^4b^3-3a^3b^4)/(6a^3b^3-2a^2b^4) ?
The solution to (9a^4b^3-3a^3b^4)/(6a^3b^3-2a^2b^4) is (3a)/2