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Trig Identity (1 Viewer)

Timothy.Siu

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nope, the way u were trying doing it was the way i did.

note that (x+y)3=x3+3x2y+3xy2+y3
this is the part u got wrong.

however if u look at drogonski's method,
he used the factorisation of cubes,
x3+y3=(x+y)(x2+xy+y2)

this shows two different ways of manipulating x3+y3 of which the latter was better for this question.
 

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