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Posted

set it equal to zero and solve (if by hand look up and use cardano's method)

 

then plug the roots into the expression (A-a1)(A-a2)(A-a3)

 

where a1,a2,a3 are the roots and should be functions of B

 

^note in the above your using A as your x.

Posted

[math]a^3 + b^3 + a^2 - b^2[/math]

Hmm... break it up into a difference of two squares and a sum of two cubes:

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

 

I think it can be simplified further from there, but I won't do the rest of your work for you.

 

 

edit: I seem to be outclassed.

Posted

so then the rest is factor by grouping........ thanks, i just wanted to check my answer because it wasnt in the back of the book and i have a test tomorrow.

Posted

not outclassed, I can't spot tricks like that for anything so I just figure out ways of guaranteeing an answer no matter what the polynomial is.

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