factorise
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a^3 - b^3 + 1 + 3ab
= (a)^3 + (-b)^3 + (1)^3 - 3(a) (-b) (1)
= ( a - b + 1) [ a^2 + b^2 + 1 - (a) (-b) - (-b) (1) - (1)(a)]
= (a-b + 1) (a^2 + b^2 + 1 + ab + b - a)
= (a)^3 + (-b)^3 + (1)^3 - 3(a) (-b) (1)
= ( a - b + 1) [ a^2 + b^2 + 1 - (a) (-b) - (-b) (1) - (1)(a)]
= (a-b + 1) (a^2 + b^2 + 1 + ab + b - a)
shikhar40:
thanks bro
ur most welcome
:-)
yo
bhaiya solve my 1 ques
go in filters and change it to primary and only maths my ques will be on first no.
Answered by
0
a^3 - b^3 + 1 + 3ab
= (a)^3 + (-b)^3 + (1)^3 - 3(a) (-b) (1)
= ( a - b + 1) [ a^2 + b^2 + 1 - (a) (-b) - (-b) (1) - (1)(a)]
= (a-b + 1) (a^2 + b^2 + 1 + ab + b - a)
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