What is 3(ab+b^2)-4(b^2-ab)+2(ba-b^2) ?

The solution to 3(ab+b^2)-4(b^2-ab)+2(ba-b^2) is 9ab-3b^2

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simplify 3(ab+b^2)-4(b^2-ab)+2(ba-b^2)

simplify

3 (ab + b^{2}) - 4 (b^{2} - ab) + 2 (ba - b^{2})

9 ab - 3 b^{2}

Solution steps

3 (ab + b^{2}) - 4 (b^{2} - ab) + 2 (ba - b^{2})

Expand

3 (ab + b^{2}) : 3 ab + 3 b^{2}

= 3 ab + 3 b^{2} - 4 (b^{2} - ab) + 2 (ba - b^{2})

Expand

- 4 (b^{2} - ab) : - 4 b^{2} + 4 ab

= 3 ab + 3 b^{2} - 4 b^{2} + 4 ab + 2 (ba - b^{2})

Expand

2 (ba - b^{2}) : 2 ab - 2 b^{2}

= 3 ab + 3 b^{2} - 4 b^{2} + 4 ab + 2 ab - 2 b^{2}

Group like terms

= 3 ab + 4 ab + 2 ab + 3 b^{2} - 4 b^{2} - 2 b^{2}

Add similar elements:

3 ab + 4 ab + 2 ab = 9 ab

= 9 ab + 3 b^{2} - 4 b^{2} - 2 b^{2}

Add similar elements:

3 b^{2} - 4 b^{2} - 2 b^{2} = - 3 b^{2}

= 9 ab - 3 b^{2}

2 i^{8} - 7 i^{9} + 2 i^{95} - 23 i^{82}

\frac{x ^{2} - 16}{2 x + 8} \cdot \frac{x ^{3} - 2 x ^{2} + x}{x ^{2} + 3 x - 4}

(\frac{1}{2} + \frac{i}{2})^{2021}

(3 x + 2) (2 x + 5)

\frac{121}{25}

What is 3(ab+b^2)-4(b^2-ab)+2(ba-b^2) ?
The solution to 3(ab+b^2)-4(b^2-ab)+2(ba-b^2) is 9ab-3b^2