Question

a^3+b^3=(a-b) (a^2+ab+b^2)​

Answer 1

Answer:

Very well known mathematical identity is

[math](a+b)^3=a^3+3a^2b+3ab^2+b^3[/math]

[math]&[/math]

[math](a-b)^3=a^3-3a^2b+3ab^2-b^3[/math]

The above can be derived easily by successive multiplication like

([math]a+b)^2=(a+b)*(a+b)[/math]

[math]=(a^2+a*b+b*a+b^2)[/math]

[math]=(a^2+2ab+b^2)[/math]

[math]So (a+b)^3[/math]

[math]=(a+b)^2*(a+b)[/math]

[math]=(a^2+2ab+b^2)*(a+b)[/math]

[math]=a^3+2a^2b+b^2a+a^2b+2ab^2+b^3[/math]

[math]=a^3+3a^2b+3ab^2+b^3[/math]

Similar for the other .

Now starting with the second identity

[math](a-b)^3=a^3-3a^2b+3ab^2-b^3[/math]

[math]Or, (a-b)^3 +3a^2b-3ab^2=a^3-b^3[/math]

[math]Or, (a-b)^3 +3ab(a-b)=a^3-b^3[/math]

[math]Or, (a-b)[ (a-b)^2 + 3ab ]=a^3-b^3[/math]

[math]Or, (a-b)(a^2–2ab+b^2+3ab)=a^3-b^3[/math]

[math]Or, (a-b)(a^2+ab+b^2)=a^3-b^3[/math]

[math]Or, a^3-b^3= (a-b)(a^2+ab+b^2)[/math]

You can do it from the opposite part like

[math](a-b)(a^2+ab+b^2)[/math]

[math]=a*(a^2+ab+b^2)-b*(a^2+ab+b^2)[/math]

[math]=a^3+a^2b+ab^2-ba^2-ab^2-b^3[/math]

[math]=a^3-b^3[/math]

So

[math]\boxed{a^3-b^3= (a-b)(a^2+ab+b^2)}[/math]

Step-by-step explanation:

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