What is (a^2b-a^2c-b^3+b^2c)/(ab-b^2-ac+bc) ?

The solution to (a^2b-a^2c-b^3+b^2c)/(ab-b^2-ac+bc) is a+b

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simplify (a^2b-a^2c-b^3+b^2c)/(ab-b^2-ac+bc)

simplify

\frac{a ^{2} b - a ^{2} c - b ^{3} + b ^{2} c}{ab - b ^{2} - a c + b c}

a + b

Solution steps

\frac{a ^{2} b - a ^{2} c - b ^{3} + b ^{2} c}{ab - b ^{2} - a c + b c}

Factor

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

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

Factor

ab - b^{2} - a c + b c : (b - c) (a - b)

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

Cancel the common factor:

b - c

= \frac{a ^{2} - b ^{2}}{a - b}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

a^{2} - b^{2} = (a + b) (a - b)

= \frac{( a + b ) ( a - b )}{a - b}

Cancel the common factor:

a - b

= a + b

(3 m - 4 n + 5) (- 3 m - 4 n)

\frac{6}{3 - 2}

2 3^{3}

\frac{36}{30}

(x + 6) (3 x - 2)

What is (a^2b-a^2c-b^3+b^2c)/(ab-b^2-ac+bc) ?
The solution to (a^2b-a^2c-b^3+b^2c)/(ab-b^2-ac+bc) is a+b