please help me please......
Attachments:
Answers
Answered by
0
if a+b+c=0 then a^3+b^3+c^3 = 3abc
= 3(a^2-b^2)+ (b^2-c^2) + (c^2-a^2)/3(a+b) + (b+c) + (c+a)
= 3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)/3(a+b)(b+c)(c+a)
= (a-b)(b-c)(c-a)
= 3(a^2-b^2)+ (b^2-c^2) + (c^2-a^2)/3(a+b) + (b+c) + (c+a)
= 3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)/3(a+b)(b+c)(c+a)
= (a-b)(b-c)(c-a)
sanidhyasingla:
why did you cut the cube
how
Its formula bro
Numerator and denominator are in the form of a^3+b^3+c^3.We know that a^3+b^3+c^3 = 3abc.
a+b+c=0 is not given
ok i understand
We have to assume that
Similar questions